Calculus Examples

$\int_{1}^{2} x^{4} \ln ({(x)}^{2}) d x$

Step 1

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$2 \int_{1}^{2} x^{4} (\ln (x)) d x$

Step 2

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (x)$ and $d v = x^{4}$ .

$2 (\ln (x) (\frac{1}{5} x^{5})]_{1}^{2} - \int_{1}^{2} \frac{1}{5} x^{5} \frac{1}{x} d x)$

Step 3

Simplify.

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Combine $\frac{1}{5}$ and $x^{5}$ .

$2 (\ln (x) \frac{x^{5}}{5}]_{1}^{2} - \int_{1}^{2} \frac{1}{5} x^{5} \frac{1}{x} d x)$

Combine $\ln (x)$ and $\frac{x^{5}}{5}$ .

$2 (\frac{\ln (x) x^{5}}{5}]_{1}^{2} - \int_{1}^{2} \frac{1}{5} x^{5} \frac{1}{x} d x)$

Combine $\frac{1}{5}$ and $x^{5}$ .

$2 (\frac{\ln (x) x^{5}}{5}]_{1}^{2} - \int_{1}^{2} \frac{x^{5}}{5} \cdot \frac{1}{x} d x)$

Multiply $\frac{x^{5}}{5}$ by $\frac{1}{x}$ .

$2 (\frac{\ln (x) x^{5}}{5}]_{1}^{2} - \int_{1}^{2} \frac{x^{5}}{5 x} d x)$

Cancel the common factor of $x^{5}$ and $x$ .

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Factor $x$ out of $x^{5}$ .

$2 (\frac{\ln (x) x^{5}}{5}]_{1}^{2} - \int_{1}^{2} \frac{x \cdot x^{4}}{5 x} d x)$

Cancel the common factors.

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Factor $x$ out of $5 x$ .

$2 (\frac{\ln (x) x^{5}}{5}]_{1}^{2} - \int_{1}^{2} \frac{x \cdot x^{4}}{x \cdot 5} d x)$

Cancel the common factor.

$2 (\frac{\ln (x) x^{5}}{5}]_{1}^{2} - \int_{1}^{2} \frac{x \cdot x^{4}}{x \cdot 5} d x)$

Rewrite the expression.

$2 (\frac{\ln (x) x^{5}}{5}]_{1}^{2} - \int_{1}^{2} \frac{x^{4}}{5} d x)$

Step 4

Since $\frac{1}{5}$ is constant with respect to $x$ , move $\frac{1}{5}$ out of the integral.

$2 (\frac{\ln (x) x^{5}}{5}]_{1}^{2} - (\frac{1}{5} \int_{1}^{2} x^{4} d x))$

Step 5

By the Power Rule, the integral of $x^{4}$ with respect to $x$ is $\frac{1}{5} x^{5}$ .

$2 (\frac{\ln (x) x^{5}}{5}]_{1}^{2} - \frac{1}{5} \frac{1}{5} x^{5}]_{1}^{2})$

Step 6

Simplify the answer.

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Simplify.

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Combine $\frac{1}{5}$ and $x^{5}$ .

$2 (\frac{\ln (x) x^{5}}{5}]_{1}^{2} - \frac{1}{5} \frac{x^{5}}{5}]_{1}^{2})$

Combine $\frac{x^{5}}{5}]_{1}^{2}$ and $\frac{1}{5}$ .

$2 (\frac{\ln (x) x^{5}}{5}]_{1}^{2} - \frac{\frac{x^{5}}{5}]_{1}^{2}}{5})$

Substitute and simplify.

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Evaluate $\frac{\ln (x) x^{5}}{5}$ at $2$ and at $1$ .

$2 ((\frac{\ln (2) \cdot 2^{5}}{5}) - \frac{\ln (1) \cdot 1^{5}}{5} - \frac{\frac{x^{5}}{5}]_{1}^{2}}{5})$

Evaluate $\frac{x^{5}}{5}$ at $2$ and at $1$ .

$2 ((\frac{\ln (2) \cdot 2^{5}}{5}) - \frac{\ln (1) \cdot 1^{5}}{5} - \frac{(\frac{2^{5}}{5}) - \frac{1^{5}}{5}}{5})$

Simplify.

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Raise $2$ to the power of $5$ .

$2 (\frac{\ln (2) \cdot 32}{5} - \frac{\ln (1) \cdot 1^{5}}{5} - \frac{(\frac{2^{5}}{5}) - \frac{1^{5}}{5}}{5})$

Move $32$ to the left of $\ln (2)$ .

$2 (\frac{32 \cdot \ln (2)}{5} - \frac{\ln (1) \cdot 1^{5}}{5} - \frac{(\frac{2^{5}}{5}) - \frac{1^{5}}{5}}{5})$

One to any power is one.

$2 (\frac{32 \ln (2)}{5} - \frac{\ln (1) \cdot 1}{5} - \frac{(\frac{2^{5}}{5}) - \frac{1^{5}}{5}}{5})$

Multiply $\ln (1)$ by $1$ .

$2 (\frac{32 \ln (2)}{5} - \frac{\ln (1)}{5} - \frac{(\frac{2^{5}}{5}) - \frac{1^{5}}{5}}{5})$

Raise $2$ to the power of $5$ .

$2 (\frac{32 \ln (2)}{5} - \frac{\ln (1)}{5} - \frac{\frac{32}{5} - \frac{1^{5}}{5}}{5})$

One to any power is one.

$2 (\frac{32 \ln (2)}{5} - \frac{\ln (1)}{5} - \frac{\frac{32}{5} - \frac{1}{5}}{5})$

Combine the numerators over the common denominator.

$2 (\frac{32 \ln (2)}{5} - \frac{\ln (1)}{5} - \frac{\frac{32 - 1}{5}}{5})$

Subtract $1$ from $32$ .

$2 (\frac{32 \ln (2)}{5} - \frac{\ln (1)}{5} - \frac{\frac{31}{5}}{5})$

Rewrite $\frac{\frac{31}{5}}{5}$ as a product.

$2 (\frac{32 \ln (2)}{5} - \frac{\ln (1)}{5} - (\frac{31}{5} \cdot \frac{1}{5}))$

Multiply $\frac{31}{5}$ by $\frac{1}{5}$ .

$2 (\frac{32 \ln (2)}{5} - \frac{\ln (1)}{5} - \frac{31}{5 \cdot 5})$

Multiply $5$ by $5$ .

$2 (\frac{32 \ln (2)}{5} - \frac{\ln (1)}{5} - \frac{31}{25})$

Step 7

Evaluate.

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Combine the numerators over the common denominator.

$2 (\frac{32 \ln (2) - \ln (1)}{5} + \frac{- 31}{25})$

Simplify each term.

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The natural logarithm of $1$ is $0$ .

$2 (\frac{32 \ln (2) - 0}{5} + \frac{- 31}{25})$

Multiply $- 1$ by $0$ .

$2 (\frac{32 \ln (2) + 0}{5} + \frac{- 31}{25})$

Add $32 \ln (2)$ and $0$ .

$2 (\frac{32 \ln (2)}{5} + \frac{- 31}{25})$

Move the negative in front of the fraction.

$2 (\frac{32 \ln (2)}{5} - \frac{31}{25})$

Apply the distributive property.

$2 \frac{32 \ln (2)}{5} + 2 (- \frac{31}{25})$

Multiply $2 \frac{32 \ln (2)}{5}$ .

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Combine $2$ and $\frac{32 \ln (2)}{5}$ .

$\frac{2 (32 \ln (2))}{5} + 2 (- \frac{31}{25})$

Multiply $32$ by $2$ .

$\frac{64 \ln (2)}{5} + 2 (- \frac{31}{25})$

Multiply $2 (- \frac{31}{25})$ .

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Multiply $- 1$ by $2$ .

$\frac{64 \ln (2)}{5} - 2 (\frac{31}{25})$

Combine $- 2$ and $\frac{31}{25}$ .

$\frac{64 \ln (2)}{5} + \frac{- 2 \cdot 31}{25}$

Multiply $- 2$ by $31$ .

$\frac{64 \ln (2)}{5} + \frac{- 62}{25}$

Move the negative in front of the fraction.

$\frac{64 \ln (2)}{5} - \frac{62}{25}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\frac{64 \ln (2)}{5} - \frac{62}{25}$

Decimal Form:

$6.39228391 \dots$