Calculus Examples

\int \frac{\ln (x)}{x^{2}} d x

$\int \frac{\ln (x)}{x^{2}} d x$

Step 1

Apply basic rules of exponents.

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Step 1.1

Move

x^{2}

$x^{2}$ out of the denominator by raising it to the

- 1

$- 1$ power.

\int \ln (x) {(x^{2})}^{- 1} d x

$\int \ln (x) {(x^{2})}^{- 1} d x$

Step 1.2

Multiply the exponents in

{(x^{2})}^{- 1}

${(x^{2})}^{- 1}$ .

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Step 1.2.1

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

\int \ln (x) x^{2 \cdot - 1} d x

$\int \ln (x) x^{2 \cdot - 1} d x$

Step 1.2.2

Multiply

2

$2$ by

- 1

$- 1$ .

\int \ln (x) x^{- 2} d x

$\int \ln (x) x^{- 2} d x$

\int \ln (x) x^{- 2} d x

$\int \ln (x) x^{- 2} d x$

\int \ln (x) x^{- 2} d x

$\int \ln (x) x^{- 2} d x$

Step 2

Integrate by parts using the formula

\int u d v = u v - \int v d u

$\int u d v = u v - \int v d u$ , where

u = \ln (x)

$u = \ln (x)$ and

d v = x^{- 2}

$d v = x^{- 2}$ .

\ln (x) (- \frac{1}{x}) - \int - \frac{1}{x} \cdot \frac{1}{x} d x

$\ln (x) (- \frac{1}{x}) - \int - \frac{1}{x} \cdot \frac{1}{x} d x$

Step 3

Simplify.

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Step 3.1

Combine

\ln (x)

$\ln (x)$ and

\frac{1}{x}

$\frac{1}{x}$ .

- \frac{\ln (x)}{x} - \int - \frac{1}{x} \cdot \frac{1}{x} d x

$- \frac{\ln (x)}{x} - \int - \frac{1}{x} \cdot \frac{1}{x} d x$

Step 3.2

Multiply

\frac{1}{x}

$\frac{1}{x}$ by

\frac{1}{x}

$\frac{1}{x}$ .

- \frac{\ln (x)}{x} - \int - \frac{1}{x \cdot x} d x

$- \frac{\ln (x)}{x} - \int - \frac{1}{x \cdot x} d x$

Step 3.3

Raise

x

$x$ to the power of

1

$1$ .

- \frac{\ln (x)}{x} - \int - \frac{1}{x^{1} x} d x

$- \frac{\ln (x)}{x} - \int - \frac{1}{x^{1} x} d x$

Step 3.4

Raise

x

$x$ to the power of

1

$1$ .

- \frac{\ln (x)}{x} - \int - \frac{1}{x^{1} x^{1}} d x

$- \frac{\ln (x)}{x} - \int - \frac{1}{x^{1} x^{1}} d x$

Step 3.5

Use the power rule

a^{m} a^{n} = a^{m + n}

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

- \frac{\ln (x)}{x} - \int - \frac{1}{x^{1 + 1}} d x

$- \frac{\ln (x)}{x} - \int - \frac{1}{x^{1 + 1}} d x$

Step 3.6

Add

1

$1$ and

1

$1$ .

- \frac{\ln (x)}{x} - \int - \frac{1}{x^{2}} d x

$- \frac{\ln (x)}{x} - \int - \frac{1}{x^{2}} d x$

- \frac{\ln (x)}{x} - \int - \frac{1}{x^{2}} d x

$- \frac{\ln (x)}{x} - \int - \frac{1}{x^{2}} d x$

Step 4

Since

- 1

$- 1$ is constant with respect to

x

$x$ , move

- 1

$- 1$ out of the integral.

- \frac{\ln (x)}{x} - - \int \frac{1}{x^{2}} d x

$- \frac{\ln (x)}{x} - - \int \frac{1}{x^{2}} d x$

Step 5

Simplify the expression.

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Step 5.1

Simplify.

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Step 5.1.1

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

- \frac{\ln (x)}{x} + 1 \int \frac{1}{x^{2}} d x

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

Step 5.1.2

Multiply

\int \frac{1}{x^{2}} d x

$\int \frac{1}{x^{2}} d x$ by

1

$1$ .

- \frac{\ln (x)}{x} + \int \frac{1}{x^{2}} d x

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

- \frac{\ln (x)}{x} + \int \frac{1}{x^{2}} d x

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

Step 5.2

Apply basic rules of exponents.

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Step 5.2.1

Move

x^{2}

$x^{2}$ out of the denominator by raising it to the

- 1

$- 1$ power.

- \frac{\ln (x)}{x} + \int {(x^{2})}^{- 1} d x

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

Step 5.2.2

Multiply the exponents in

{(x^{2})}^{- 1}

${(x^{2})}^{- 1}$ .

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Step 5.2.2.1

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

- \frac{\ln (x)}{x} + \int x^{2 \cdot - 1} d x

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

Step 5.2.2.2

Multiply

2

$2$ by

- 1

$- 1$ .

- \frac{\ln (x)}{x} + \int x^{- 2} d x

$- \frac{\ln (x)}{x} + \int x^{- 2} d x$

- \frac{\ln (x)}{x} + \int x^{- 2} d x

$- \frac{\ln (x)}{x} + \int x^{- 2} d x$

- \frac{\ln (x)}{x} + \int x^{- 2} d x

$- \frac{\ln (x)}{x} + \int x^{- 2} d x$

- \frac{\ln (x)}{x} + \int x^{- 2} d x

$- \frac{\ln (x)}{x} + \int x^{- 2} d x$

Step 6

By the Power Rule, the integral of

x^{- 2}

$x^{- 2}$ with respect to

x

$x$ is

- x^{- 1}

$- x^{- 1}$ .

- \frac{\ln (x)}{x} - x^{- 1} + C

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

Step 7

Rewrite

- \frac{\ln (x)}{x} - x^{- 1} + C

$- \frac{\ln (x)}{x} - x^{- 1} + C$ as

- \frac{\ln (x)}{x} - \frac{1}{x} + C

$- \frac{\ln (x)}{x} - \frac{1}{x} + C$ .

- \frac{\ln (x)}{x} - \frac{1}{x} + C

$- \frac{\ln (x)}{x} - \frac{1}{x} + C$