Calculus Examples

$\frac{1}{x} \cdot \frac{d y}{d x} - \frac{1}{1 + x^{2}} y = x^{3}$

Step 1

Rewrite the differential equation as $\frac{d y}{d x} + M (x) y = Q (x)$ .

Tap for more steps...

Step 1.1

Rewrite the equation as $M (x) \frac{d y}{d x} + P (x) y = Q (x)$ .

Tap for more steps...

Step 1.1.1

Reorder terms.

$\frac{1}{x} \frac{d y}{d x} - \frac{1}{x^{2} + 1} y = x^{3}$

Step 1.1.2

Reorder terms.

$\frac{1}{x} \frac{d y}{d x} - y \frac{1}{x^{2} + 1} = x^{3}$

Step 1.2

Factor $y$ out of $- y \frac{1}{x^{2} + 1}$ .

$\frac{1}{x} \frac{d y}{d x} + y (- \frac{1}{x^{2} + 1}) = x^{3}$

Step 1.3

Reorder $y$ and $- \frac{1}{x^{2} + 1}$ .

$\frac{1}{x} \frac{d y}{d x} - \frac{1}{x^{2} + 1} y = x^{3}$

Step 1.4

Multiply each term in $\frac{1}{x} \frac{d y}{d x} - \frac{1}{x^{2} + 1} y = x^{3}$ by $x$ .

$\frac{1}{x} \frac{d y}{d x} x - \frac{1}{x^{2} + 1} y x = x^{3} x$

Step 1.5

Combine $\frac{1}{x}$ and $\frac{d y}{d x}$ .

$\frac{\frac{d y}{d x}}{x} x - \frac{1}{x^{2} + 1} y x = x^{3} x$

Step 1.6

Combine $\frac{\frac{d y}{d x}}{x}$ and $x$ .

$\frac{\frac{d y}{d x} x}{x} - \frac{1}{x^{2} + 1} y x = x^{3} x$

Step 1.7

Combine $y$ and $\frac{1}{x^{2} + 1}$ .

$\frac{\frac{d y}{d x} x}{x} - \frac{y}{x^{2} + 1} x = x^{3} x$

Step 1.8

Combine $x$ and $\frac{y}{x^{2} + 1}$ .

$\frac{\frac{d y}{d x} x}{x} - \frac{x y}{x^{2} + 1} = x^{3} x$

Step 1.9

Multiply $x^{3}$ by $x$ by adding the exponents.

Tap for more steps...

Step 1.9.1

Multiply $x^{3}$ by $x$ .

Tap for more steps...

Step 1.9.1.1

Raise $x$ to the power of $1$ .

$\frac{\frac{d y}{d x} x}{x} - \frac{x y}{x^{2} + 1} = x^{3} x^{1}$

Step 1.9.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\frac{d y}{d x} x}{x} - \frac{x y}{x^{2} + 1} = x^{3 + 1}$

Step 1.9.2

Add $3$ and $1$ .

$\frac{\frac{d y}{d x} x}{x} - \frac{x y}{x^{2} + 1} = x^{4}$

Step 1.10

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.10.1

Cancel the common factor.

$\frac{\frac{d y}{d x} x}{x} - \frac{x y}{x^{2} + 1} = x^{4}$

Step 1.10.2

Divide $\frac{d y}{d x}$ by $1$ .

$\frac{d y}{d x} - \frac{x y}{x^{2} + 1} = x^{4}$

Step 1.11

Factor $y$ out of $- \frac{x y}{x^{2} + 1}$ .

$\frac{d y}{d x} + y (- \frac{x}{x^{2} + 1}) = x^{4}$

Step 1.12

Reorder $y$ and $- \frac{x}{x^{2} + 1}$ .

$\frac{d y}{d x} - \frac{x}{x^{2} + 1} y = x^{4}$

Step 2

The integrating factor is defined by the formula $e^{\int P (x) d x}$ , where $P (x) = - \frac{x}{x^{2} + 1}$ .

Tap for more steps...

Step 2.1

Set up the integration.

$e^{\int - \frac{x}{x^{2} + 1} d x}$

Step 2.2

Integrate $- \frac{x}{x^{2} + 1}$ .

Tap for more steps...

Step 2.2.1

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$e^{- \int \frac{x}{x^{2} + 1} d x}$

Step 2.2.2

Let $u = x^{2} + 1$ . Then $d u = 2 x d x$ , so $\frac{1}{2} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

Tap for more steps...

Step 2.2.2.1

Let $u = x^{2} + 1$ . Find $\frac{d u}{d x}$ .

Tap for more steps...

Step 2.2.2.1.1

Differentiate $x^{2} + 1$ .

$\frac{d}{d x} [x^{2} + 1]$

Step 2.2.2.1.2

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$

Step 2.2.2.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$2 x + \frac{d}{d x} [1]$

Step 2.2.2.1.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$2 x + 0$

Step 2.2.2.1.5

Add $2 x$ and $0$ .

$2 x$

Step 2.2.2.2

Rewrite the problem using $u$ and $d u$ .

$e^{- \int \frac{1}{u} \cdot \frac{1}{2} d u}$

Step 2.2.3

Simplify.

Tap for more steps...

Step 2.2.3.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$e^{- \int \frac{1}{u \cdot 2} d u}$

Step 2.2.3.2

Move $2$ to the left of $u$ .

$e^{- \int \frac{1}{2 u} d u}$

Step 2.2.4

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$e^{- (\frac{1}{2} \int \frac{1}{u} d u)}$

Step 2.2.5

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$e^{- \frac{1}{2} (\ln (| u |) + C)}$

Step 2.2.6

Simplify.

$e^{- \frac{1}{2} \ln (| u |) + C}$

Step 2.2.7

Replace all occurrences of $u$ with $x^{2} + 1$ .

$e^{- \frac{1}{2} \ln (| x^{2} + 1 |) + C}$

Step 2.3

Remove the constant of integration.

$e^{- \frac{1}{2} \ln (x^{2} + 1)}$

Step 2.4

Use the logarithmic power rule.

$e^{\ln ({(x^{2} + 1)}^{- \frac{1}{2}})}$

Step 2.5

Exponentiation and log are inverse functions.

${(x^{2} + 1)}^{- \frac{1}{2}}$

Step 2.6

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}}$

Step 3

Multiply each term by the integrating factor $\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}}$ .

Tap for more steps...

Step 3.1

Multiply each term by $\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}}$ .

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} \frac{d y}{d x} + \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} (- \frac{x}{x^{2} + 1} y) = \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} x^{4}$

Step 3.2

Simplify each term.

Tap for more steps...

Step 3.2.1

Combine $\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}}$ and $\frac{d y}{d x}$ .

$\frac{\frac{d y}{d x}}{{(x^{2} + 1)}^{\frac{1}{2}}} + \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} (- \frac{x}{x^{2} + 1} y) = \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} x^{4}$

Step 3.2.2

Rewrite using the commutative property of multiplication.

$\frac{\frac{d y}{d x}}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} (\frac{x}{x^{2} + 1} y) = \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} x^{4}$

Step 3.2.3

Combine $\frac{x}{x^{2} + 1}$ and $y$ .

$\frac{\frac{d y}{d x}}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} \cdot \frac{x y}{x^{2} + 1} = \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} x^{4}$

Step 3.2.4

Multiply $- \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} \cdot \frac{x y}{x^{2} + 1}$ .

Tap for more steps...

Step 3.2.4.1

Multiply $\frac{x y}{x^{2} + 1}$ by $\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}}$ .

$\frac{\frac{d y}{d x}}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{x y}{(x^{2} + 1) {(x^{2} + 1)}^{\frac{1}{2}}} = \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} x^{4}$

Step 3.2.4.2

Multiply $x^{2} + 1$ by ${(x^{2} + 1)}^{\frac{1}{2}}$ by adding the exponents.

Tap for more steps...

Step 3.2.4.2.1

Multiply $x^{2} + 1$ by ${(x^{2} + 1)}^{\frac{1}{2}}$ .

Tap for more steps...

Step 3.2.4.2.1.1

Raise $x^{2} + 1$ to the power of $1$ .

$\frac{\frac{d y}{d x}}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{x y}{{(x^{2} + 1)}^{1} {(x^{2} + 1)}^{\frac{1}{2}}} = \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} x^{4}$

Step 3.2.4.2.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\frac{d y}{d x}}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{x y}{{(x^{2} + 1)}^{1 + \frac{1}{2}}} = \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} x^{4}$

Step 3.2.4.2.2

Write $1$ as a fraction with a common denominator.

$\frac{\frac{d y}{d x}}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{x y}{{(x^{2} + 1)}^{\frac{2}{2} + \frac{1}{2}}} = \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} x^{4}$

Step 3.2.4.2.3

Combine the numerators over the common denominator.

$\frac{\frac{d y}{d x}}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{x y}{{(x^{2} + 1)}^{\frac{2 + 1}{2}}} = \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} x^{4}$

Step 3.2.4.2.4

Add $2$ and $1$ .

$\frac{\frac{d y}{d x}}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{x y}{{(x^{2} + 1)}^{\frac{3}{2}}} = \frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} x^{4}$

Step 3.3

Combine $\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}}$ and $x^{4}$ .

$\frac{\frac{d y}{d x}}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{x y}{{(x^{2} + 1)}^{\frac{3}{2}}} = \frac{x^{4}}{{(x^{2} + 1)}^{\frac{1}{2}}}$

Step 4

Rewrite the left side as a result of differentiating a product.

$\frac{d}{d x} [\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y] = \frac{x^{4}}{{(x^{2} + 1)}^{\frac{1}{2}}}$

Step 5

Set up an integral on each side.

$\int \frac{d}{d x} [\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y] d x = \int \frac{x^{4}}{{(x^{2} + 1)}^{\frac{1}{2}}} d x$

Step 6

Integrate the left side.

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int \frac{x^{4}}{{(x^{2} + 1)}^{\frac{1}{2}}} d x$

Step 7

Integrate the right side.

Tap for more steps...

Step 7.1

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int \frac{x^{4}}{\sqrt{{(x^{2} + 1)}^{1}}} d x$

Step 7.2

Let $x = \tan (t_{1})$ , where $- \frac{π}{2} \leq t_{1} \leq \frac{π}{2}$ . Then $d x = \sec^{2} (t_{1}) d t_{1}$ . Note that since $- \frac{π}{2} \leq t_{1} \leq \frac{π}{2}$ , $\sec^{2} (t_{1})$ is positive.

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int \frac{\tan^{4} (t_{1})}{\sqrt{{(\tan^{2} (t_{1}) + 1)}^{1}}} \sec^{2} (t_{1}) d t_{1}$

Step 7.3

Simplify terms.

Tap for more steps...

Step 7.3.1

Simplify $\sqrt{{(\tan^{2} (t_{1}) + 1)}^{1}}$ .

Tap for more steps...

Step 7.3.1.1

Apply pythagorean identity.

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int \frac{\tan^{4} (t_{1})}{\sqrt{{(\sec^{2} (t_{1}))}^{1}}} \sec^{2} (t_{1}) d t_{1}$

Step 7.3.1.2

Multiply the exponents in ${(\sec^{2} (t_{1}))}^{1}$ .

Tap for more steps...

Step 7.3.1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int \frac{\tan^{4} (t_{1})}{\sqrt{{\sec (t_{1})}^{2 \cdot 1}}} \sec^{2} (t_{1}) d t_{1}$

Step 7.3.1.2.2

Multiply $2$ by $1$ .

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int \frac{\tan^{4} (t_{1})}{\sqrt{\sec^{2} (t_{1})}} \sec^{2} (t_{1}) d t_{1}$

Step 7.3.1.3

Pull terms out from under the radical, assuming positive real numbers.

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int \frac{\tan^{4} (t_{1})}{\sec (t_{1})} \sec^{2} (t_{1}) d t_{1}$

Step 7.3.2

Cancel the common factor of $\sec (t_{1})$ .

Tap for more steps...

Step 7.3.2.1

Factor $\sec (t_{1})$ out of $\sec^{2} (t_{1})$ .

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int \frac{\tan^{4} (t_{1})}{\sec (t_{1})} (\sec (t_{1}) \sec (t_{1})) d t_{1}$

Step 7.3.2.2

Cancel the common factor.

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int \frac{\tan^{4} (t_{1})}{\sec (t_{1})} (\sec (t_{1}) \sec (t_{1})) d t_{1}$

Step 7.3.2.3

Rewrite the expression.

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int \tan^{4} (t_{1}) \sec (t_{1}) d t_{1}$

Step 7.4

Raise $\sec (t_{1})$ to the power of $1$ .

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int \tan^{4} (t_{1}) \sec^{1} (t_{1}) d t_{1}$

Step 7.5

Simplify with factoring out.

Tap for more steps...

Step 7.5.1

Factor $2$ out of $4$ .

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int {\tan (t_{1})}^{2 (2)} \sec^{1} (t_{1}) d t_{1}$

Step 7.5.2

Rewrite ${\tan (t_{1})}^{2 (2)}$ as exponentiation.

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int {(\tan^{2} (t_{1}))}^{2} \sec^{1} (t_{1}) d t_{1}$

Step 7.6

Using the Pythagorean Identity, rewrite $\tan^{2} (t_{1})$ as $- 1 + \sec^{2} (t_{1})$ .

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int {(- 1 + \sec^{2} (t_{1}))}^{2} \sec^{1} (t_{1}) d t_{1}$

Step 7.7

Simplify.

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int (1 - 2 \sec^{2} (t) + \sec^{4} (t)) \sec^{1} (t_{1}) d t_{1}$

Step 7.8

Simplify terms.

Tap for more steps...

Step 7.8.1

Apply the distributive property.

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int 1 \sec^{1} (t_{1}) - 2 \sec^{2} (t) \sec^{1} (t_{1}) + \sec^{4} (t) \sec^{1} (t_{1}) d t_{1}$

Step 7.8.2

Simplify each term.

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int \sec (t_{1}) - 2 \sec^{2} (t) \sec (t_{1}) + \sec^{4} (t) \sec (t_{1}) d t_{1}$

Step 7.9

Split the single integral into multiple integrals.

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \int \sec (t_{1}) d t_{1} + \int - 2 \sec^{2} (t) \sec (t_{1}) d t_{1} + \int \sec^{4} (t) \sec (t_{1}) d t_{1}$

Step 7.10

The integral of $\sec (t_{1})$ with respect to $t_{1}$ is $\ln (| \sec (t_{1}) + \tan (t_{1}) |)$ .

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \ln (| \sec (t_{1}) + \tan (t_{1}) |) + C + \int - 2 \sec^{2} (t) \sec (t_{1}) d t_{1} + \int \sec^{4} (t) \sec (t_{1}) d t_{1}$

Step 7.11

Since $- 2 \sec^{2} (t)$ is constant with respect to $t_{1}$ , move $- 2 \sec^{2} (t)$ out of the integral.

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \ln (| \sec (t_{1}) + \tan (t_{1}) |) + C - 2 \sec^{2} (t) \int \sec (t_{1}) d t_{1} + \int \sec^{4} (t) \sec (t_{1}) d t_{1}$

Step 7.12

The integral of $\sec (t_{1})$ with respect to $t_{1}$ is $\ln (| \sec (t_{1}) + \tan (t_{1}) |)$ .

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \ln (| \sec (t_{1}) + \tan (t_{1}) |) + C - 2 \sec^{2} (t) (\ln (| \sec (t_{1}) + \tan (t_{1}) |) + C) + \int \sec^{4} (t) \sec (t_{1}) d t_{1}$

Step 7.13

Since $\sec^{4} (t)$ is constant with respect to $t_{1}$ , move $\sec^{4} (t)$ out of the integral.

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \ln (| \sec (t_{1}) + \tan (t_{1}) |) + C - 2 \sec^{2} (t) (\ln (| \sec (t_{1}) + \tan (t_{1}) |) + C) + \sec^{4} (t) \int \sec (t_{1}) d t_{1}$

Step 7.14

The integral of $\sec (t_{1})$ with respect to $t_{1}$ is $\ln (| \sec (t_{1}) + \tan (t_{1}) |)$ .

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \ln (| \sec (t_{1}) + \tan (t_{1}) |) + C - 2 \sec^{2} (t) (\ln (| \sec (t_{1}) + \tan (t_{1}) |) + C) + \sec^{4} (t) (\ln (| \sec (t_{1}) + \tan (t_{1}) |) + C)$

Step 7.15

Simplify.

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \ln (| \sec (t_{1}) + \tan (t_{1}) |) - 2 \sec^{2} (t) \ln (| \sec (t_{1}) + \tan (t_{1}) |) + \sec^{4} (t) \ln (| \sec (t_{1}) + \tan (t_{1}) |) + C$

Step 7.16

Replace all occurrences of $t_{1}$ with $\arctan (x)$ .

$\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}} y = \ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) - 2 \sec^{2} (t) \ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) + \sec^{4} (t) \ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) + C$

Step 8

Solve for $y$ .

Tap for more steps...

Step 8.1

Simplify the left side.

Tap for more steps...

Step 8.1.1

Combine $\frac{1}{{(x^{2} + 1)}^{\frac{1}{2}}}$ and $y$ .

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) - 2 \sec^{2} (t) \ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) + \sec^{4} (t) \ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) + C$

Step 8.2

Simplify the right side.

Tap for more steps...

Step 8.2.1

Simplify each term.

Tap for more steps...

Step 8.2.1.1

Simplify each term.

Tap for more steps...

Step 8.2.1.1.1

Draw a triangle in the plane with vertices $(1, x)$ , $(1, 0)$ , and the origin. Then $\arctan (x)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(1, x)$ . Therefore, $\sec (\arctan (x))$ is $\sqrt{1 + x^{2}}$ .

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + \tan (\arctan (x)) |) - 2 \sec^{2} (t) \ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) + \sec^{4} (t) \ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) + C$

Step 8.2.1.1.2

The functions tangent and arctangent are inverses.

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + x |) - 2 \sec^{2} (t) \ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) + \sec^{4} (t) \ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) + C$

Step 8.2.1.2

Simplify each term.

Tap for more steps...

Step 8.2.1.2.1

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + x |) - 2 \sec^{2} (t) \ln (| \sqrt{1 + x^{2}} + \tan (\arctan (x)) |) + \sec^{4} (t) \ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) + C$

Step 8.2.1.2.2

The functions tangent and arctangent are inverses.

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + x |) - 2 \sec^{2} (t) \ln (| \sqrt{1 + x^{2}} + x |) + \sec^{4} (t) \ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) + C$

Step 8.2.1.3

Multiply $- 2 \sec^{2} (t) \ln (| \sqrt{1 + x^{2}} + x |)$ .

Tap for more steps...

Step 8.2.1.3.1

Reorder $\ln (| \sqrt{1 + x^{2}} + x |)$ and $- 2$ .

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + x |) - 2 \cdot \ln (| \sqrt{1 + x^{2}} + x |) \sec^{2} (t) + \sec^{4} (t) \ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) + C$

Step 8.2.1.3.2

Simplify $- 2 \cdot \ln (| \sqrt{1 + x^{2}} + x |)$ by moving $2$ inside the logarithm.

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + x |) - \ln ({| \sqrt{1 + x^{2}} + x |}^{2}) \sec^{2} (t) + \sec^{4} (t) \ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) + C$

Step 8.2.1.4

Remove the absolute value in ${| \sqrt{1 + x^{2}} + x |}^{2}$ because exponentiations with even powers are always positive.

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + x |) - \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) + \sec^{4} (t) \ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) + C$

Step 8.2.1.5

Simplify each term.

Tap for more steps...

Step 8.2.1.5.1

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + x |) - \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) + \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + \tan (\arctan (x)) |) + C$

Step 8.2.1.5.2

The functions tangent and arctangent are inverses.

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + x |) - \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) + \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |) + C$

Step 8.3

Move all the terms containing a logarithm to the left side of the equation.

$- \ln (| \sqrt{1 + x^{2}} + x |) + \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) - \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |) = - \frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} + C$

Step 8.4

Rewrite the equation as $- \frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} + C = - \ln (| \sqrt{1 + x^{2}} + x |) + \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) - \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |)$ .

$- \frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} + C = - \ln (| \sqrt{1 + x^{2}} + x |) + \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) - \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |)$

Step 8.5

Subtract $C$ from both sides of the equation.

$- \frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = - \ln (| \sqrt{1 + x^{2}} + x |) + \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) - \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |) - C$

Step 8.6

Divide each term in $- \frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = - \ln (| \sqrt{1 + x^{2}} + x |) + \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) - \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |) - C$ by $- 1$ and simplify.

Tap for more steps...

Step 8.6.1

$\frac{- \frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}}}{- 1} = \frac{- \ln (| \sqrt{1 + x^{2}} + x |)}{- 1} + \frac{\ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t)}{- 1} + \frac{- \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |)}{- 1} + \frac{- C}{- 1}$

Step 8.6.2

Simplify the left side.

Tap for more steps...

Step 8.6.2.1

Dividing two negative values results in a positive value.

$\frac{\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}}}{1} = \frac{- \ln (| \sqrt{1 + x^{2}} + x |)}{- 1} + \frac{\ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t)}{- 1} + \frac{- \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |)}{- 1} + \frac{- C}{- 1}$

Step 8.6.2.2

Divide $\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}}$ by $1$ .

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \frac{- \ln (| \sqrt{1 + x^{2}} + x |)}{- 1} + \frac{\ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t)}{- 1} + \frac{- \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |)}{- 1} + \frac{- C}{- 1}$

Step 8.6.3

Simplify the right side.

Tap for more steps...

Step 8.6.3.1

Simplify each term.

Tap for more steps...

Step 8.6.3.1.1

Dividing two negative values results in a positive value.

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \frac{\ln (| \sqrt{1 + x^{2}} + x |)}{1} + \frac{\ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t)}{- 1} + \frac{- \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |)}{- 1} + \frac{- C}{- 1}$

Step 8.6.3.1.2

Divide $\ln (| \sqrt{1 + x^{2}} + x |)$ by $1$ .

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + x |) + \frac{\ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t)}{- 1} + \frac{- \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |)}{- 1} + \frac{- C}{- 1}$

Step 8.6.3.1.3

Move the negative one from the denominator of $\frac{\ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t)}{- 1}$ .

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + x |) - 1 \cdot (\ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t)) + \frac{- \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |)}{- 1} + \frac{- C}{- 1}$

Step 8.6.3.1.4

Rewrite $- 1 \cdot (\ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t))$ as $- (\ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t))$ .

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + x |) - (\ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t)) + \frac{- \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |)}{- 1} + \frac{- C}{- 1}$

Step 8.6.3.1.5

Dividing two negative values results in a positive value.

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + x |) - \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) + \frac{\sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |)}{1} + \frac{- C}{- 1}$

Step 8.6.3.1.6

Divide $\sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |)$ by $1$ .

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + x |) - \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) + \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |) + \frac{- C}{- 1}$

Step 8.6.3.1.7

Dividing two negative values results in a positive value.

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + x |) - \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) + \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |) + \frac{C}{1}$

Step 8.6.3.1.8

Divide $C$ by $1$ .

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} = \ln (| \sqrt{1 + x^{2}} + x |) - \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) + \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |) + C$

Step 8.7

Multiply both sides by ${(x^{2} + 1)}^{\frac{1}{2}}$ .

$\frac{y}{{(x^{2} + 1)}^{\frac{1}{2}}} {(x^{2} + 1)}^{\frac{1}{2}} = (\ln (| \sqrt{1 + x^{2}} + x |) - \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) + \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |) + C) {(x^{2} + 1)}^{\frac{1}{2}}$

Step 8.8

Simplify.

Tap for more steps...

Step 8.8.1

Simplify the left side.

Tap for more steps...

Step 8.8.1.1

Cancel the common factor of ${(x^{2} + 1)}^{\frac{1}{2}}$ .

Tap for more steps...

Step 8.8.1.1.1

Cancel the common factor.

Step 8.8.1.1.2

Rewrite the expression.

$y = (\ln (| \sqrt{1 + x^{2}} + x |) - \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) + \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |) + C) {(x^{2} + 1)}^{\frac{1}{2}}$

Step 8.8.2

Simplify the right side.

Tap for more steps...

Step 8.8.2.1

Simplify $(\ln (| \sqrt{1 + x^{2}} + x |) - \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) + \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |) + C) {(x^{2} + 1)}^{\frac{1}{2}}$ .

Tap for more steps...

Step 8.8.2.1.1

Apply the distributive property.

$y = \ln (| \sqrt{1 + x^{2}} + x |) {(x^{2} + 1)}^{\frac{1}{2}} - \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) {(x^{2} + 1)}^{\frac{1}{2}} + \sec^{4} (t) \ln (| \sqrt{1 + x^{2}} + x |) {(x^{2} + 1)}^{\frac{1}{2}} + C {(x^{2} + 1)}^{\frac{1}{2}}$

Step 8.8.2.1.2

Simplify the expression.

Tap for more steps...

Step 8.8.2.1.2.1

Reorder $\sec^{4} (t)$ and $\ln (| \sqrt{1 + x^{2}} + x |)$ .

$y = \ln (| \sqrt{1 + x^{2}} + x |) {(x^{2} + 1)}^{\frac{1}{2}} - \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) {(x^{2} + 1)}^{\frac{1}{2}} + \ln (| \sqrt{1 + x^{2}} + x |) \sec^{4} (t) {(x^{2} + 1)}^{\frac{1}{2}} + C {(x^{2} + 1)}^{\frac{1}{2}}$

Step 8.8.2.1.2.2

Move $\ln (| \sqrt{1 + x^{2}} + x |) {(x^{2} + 1)}^{\frac{1}{2}}$ .

$y = - \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) {(x^{2} + 1)}^{\frac{1}{2}} + \ln (| \sqrt{1 + x^{2}} + x |) \sec^{4} (t) {(x^{2} + 1)}^{\frac{1}{2}} + C {(x^{2} + 1)}^{\frac{1}{2}} + \ln (| \sqrt{1 + x^{2}} + x |) {(x^{2} + 1)}^{\frac{1}{2}}$

Step 8.8.2.1.2.3

Reorder $- \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) {(x^{2} + 1)}^{\frac{1}{2}}$ and $\ln (| \sqrt{1 + x^{2}} + x |) \sec^{4} (t) {(x^{2} + 1)}^{\frac{1}{2}}$ .

$y = \ln (| \sqrt{1 + x^{2}} + x |) \sec^{4} (t) {(x^{2} + 1)}^{\frac{1}{2}} - \ln ({(\sqrt{1 + x^{2}} + x)}^{2}) \sec^{2} (t) {(x^{2} + 1)}^{\frac{1}{2}} + C {(x^{2} + 1)}^{\frac{1}{2}} + \ln (| \sqrt{1 + x^{2}} + x |) {(x^{2} + 1)}^{\frac{1}{2}}$