Calculus Examples

$\int \frac{1}{\sqrt{8 - 2 x - x^{2}}} d x$

Step 1

Rewrite $8$ as $9$ plus $- 1$

$\int \frac{1}{\sqrt{9 - 1 - 2 x - x^{2}}} d x$

Step 2

Factor using the perfect square rule.

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Step 2.1

Rewrite $1$ as $1^{2}$ .

$\int \frac{1}{\sqrt{9 - (1^{2} + 2 x + x^{2})}} d x$

Step 2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot 1 \cdot x$

Step 2.3

Rewrite the polynomial.

$\int \frac{1}{\sqrt{9 - (1^{2} + 2 \cdot 1 \cdot x + x^{2})}} d x$

Step 2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 1$ and $b = x$ .

$\int \frac{1}{\sqrt{9 - {(1 + x)}^{2}}} d x$

Step 3

Rewrite $9$ as $3^{2}$ .

$\int \frac{1}{\sqrt{3^{2} - {(1 + x)}^{2}}} d x$

Step 4

Let $u = 1 + x$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.1

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

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Step 4.1.1

Differentiate $1 + x$ .

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

Step 4.1.2

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

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

Step 4.1.3

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

$0 + \frac{d}{d x} [x]$

Step 4.1.4

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

$0 + 1$

Step 4.1.5

Add $0$ and $1$ .

$1$

Step 4.2

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

$\int \frac{1}{\sqrt{3^{2} - u^{2}}} d u$

Step 5

The integral of $\frac{1}{\sqrt{3^{2} - u^{2}}}$ with respect to $u$ is $\arcsin (\frac{u}{3})$

$\arcsin (\frac{u}{3}) + C$

Step 6

Rewrite $\arcsin (\frac{u}{3}) + C$ as $\arcsin (\frac{1}{3} u) + C$ .

$\arcsin (\frac{1}{3} u) + C$

Step 7

Replace all occurrences of $u$ with $1 + x$ .

$\arcsin (\frac{1}{3} (1 + x)) + C$