Calculus Examples

$\int_{1}^{e} \frac{1}{y (1 + \ln (y))} d y$

Step 1

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

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Step 1.1

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

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Step 1.1.1

Differentiate $\ln (y)$ .

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

Step 1.1.2

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

$\frac{1}{y}$

Step 1.2

Substitute the lower limit in for $y$ in $u_{1} = \ln (y)$ .

$u_{lower} = \ln (1)$

Step 1.3

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

$u_{lower} = 0$

Step 1.4

Substitute the upper limit in for $y$ in $u_{1} = \ln (y)$ .

$u_{upper} = \ln (e)$

Step 1.5

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

$u_{upper} = 1$

Step 1.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 0$

$u_{upper} = 1$

Step 1.7

Rewrite the problem using $u_{1}$ , $d u_{1}$ , and the new limits of integration.

$\int_{0}^{1} \frac{1}{1 + u_{1}} d u_{1}$

Step 2

Let $u_{2} = 1 + u_{1}$ . Then $d u_{2} = d u_{1}$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.1

Let $u_{2} = 1 + u_{1}$ . Find $\frac{d u_{2}}{d u_{1}}$ .

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Step 2.1.1

Differentiate $1 + u_{1}$ .

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

Step 2.1.2

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

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

Step 2.1.3

Since $1$ is constant with respect to $u_{1}$ , the derivative of $1$ with respect to $u_{1}$ is $0$ .

$0 + \frac{d}{d u_{1}} [u_{1}]$

Step 2.1.4

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

$0 + 1$

Step 2.1.5

Add $0$ and $1$ .

$1$

Step 2.2

Substitute the lower limit in for $u_{1}$ in $u_{2} = 1 + u_{1}$ .

$u_{lower} = 1 + 0$

Step 2.3

Add $1$ and $0$ .

$u_{lower} = 1$

Step 2.4

Substitute the upper limit in for $u_{1}$ in $u_{2} = 1 + u_{1}$ .

$u_{upper} = 1 + 1$

Step 2.5

Add $1$ and $1$ .

$u_{upper} = 2$

Step 2.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 1$

$u_{upper} = 2$

Step 2.7

Rewrite the problem using $u_{2}$ , $d u_{2}$ , and the new limits of integration.

$\int_{1}^{2} \frac{1}{u_{2}} d u_{2}$

Step 3

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\ln (| u_{2} |)]_{1}^{2}$

Step 4

Evaluate $\ln (| u_{2} |)$ at $2$ and at $1$ .

$\ln (| 2 |) - \ln (| 1 |)$

Step 5

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| 2 |}{| 1 |})$

Step 6

Simplify.

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Step 6.1

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\ln (\frac{2}{| 1 |})$

Step 6.2

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\ln (\frac{2}{1})$

Step 6.3

Divide $2$ by $1$ .

$\ln (2)$

Step 7

The result can be shown in multiple forms.

Exact Form:

$\ln (2)$

Decimal Form:

$0.69314718 \dots$