Calculus Examples

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

Step 1

Let $u_{1} = \ln (x)$ . Then $d u_{1} = \frac{1}{x} d x$ , so $x d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 1.1

Let $u_{1} = \ln (x)$ . Find $\frac{d u_{1}}{d x}$ .

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Step 1.1.1

Differentiate $\ln (x)$ .

$\frac{d}{d x} [\ln (x)]$

Step 1.1.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$\frac{1}{x}$

Step 1.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int \frac{u_{1}}{e^{u_{1}}} d u_{1}$

Step 2

Simplify the expression.

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Step 2.1

Negate the exponent of $e^{u_{1}}$ and move it out of the denominator.

$\int u_{1} {(e^{u_{1}})}^{- 1} d u_{1}$

Step 2.2

Multiply the exponents in ${(e^{u_{1}})}^{- 1}$ .

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Step 2.2.1

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

$\int u_{1} e^{u_{1} \cdot - 1} d u_{1}$

Step 2.2.2

Move $- 1$ to the left of $u_{1}$ .

$\int u_{1} e^{- 1 \cdot u_{1}} d u_{1}$

Step 2.2.3

Rewrite $- 1 u_{1}$ as $- u_{1}$ .

$\int u_{1} e^{- u_{1}} d u_{1}$

Step 3

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = u_{1}$ and $d v = e^{- u_{1}}$ .

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

Step 4

Apply the constant rule.

$u_{1} (- e^{u_{2}}) - (- e^{u_{2}} u_{1} + C)$

Step 5

Simplify.

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Step 5.1

Rewrite $u_{1} (- e^{u_{2}}) - (- e^{u_{2}} u_{1} + C)$ as $- u_{1} e^{u_{2}} - (- e^{u_{2}} u_{1}) + C$ .

$- u_{1} e^{u_{2}} - (- e^{u_{2}} u_{1}) + C$

Step 5.2

Rewrite $- u_{1} e^{u_{2}} - (- e^{u_{2}} u_{1}) + C$ as $- u_{1} e^{u_{2}} + e^{u_{2}} u_{1} + C$ .

$- u_{1} e^{u_{2}} + e^{u_{2}} u_{1} + C$

Step 5.3

Simplify.

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Step 5.3.1

Add $- u_{1} e^{u_{2}}$ and $e^{u_{2}} u_{1}$ .

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Step 5.3.1.1

Reorder $e^{u_{2}}$ and $u_{1}$ .

$- u_{1} e^{u_{2}} + u_{1} e^{u_{2}} + C$

Step 5.3.1.2

Add $- u_{1} e^{u_{2}}$ and $u_{1} e^{u_{2}}$ .

$0 + C$

Step 5.3.2

Add $0$ and $C$ .

$C$