Calculus Examples

$\int_{\sqrt{x}}^{2 x} \arctan (t) d t$

Step 1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \arctan (t)$ and $d v = 1$ .

$\arctan (t) t]_{\sqrt{x}}^{2 x} - \int_{\sqrt{x}}^{2 x} t \frac{1}{t^{2} + 1} d t$

Step 2

Combine $t$ and $\frac{1}{t^{2} + 1}$ .

$\arctan (t) t]_{\sqrt{x}}^{2 x} - \int_{\sqrt{x}}^{2 x} \frac{t}{t^{2} + 1} d t$

Step 3

Let $u = t^{2} + 1$ . Then $d u = 2 t d t$ , so $\frac{1}{2} d u = t d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 3.1

Let $u = t^{2} + 1$ . Find $\frac{d u}{d t}$ .

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Step 3.1.1

Differentiate $t^{2} + 1$ .

$\frac{d}{d t} [t^{2} + 1]$

Step 3.1.2

By the Sum Rule, the derivative of $t^{2} + 1$ with respect to $t$ is $\frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$ .

$\frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$

Step 3.1.3

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$2 t + \frac{d}{d t} [1]$

Step 3.1.4

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

$2 t + 0$

Step 3.1.5

Add $2 t$ and $0$ .

$2 t$

Step 3.2

Substitute the lower limit in for $t$ in $u = t^{2} + 1$ .

$u_{lower} = {\sqrt{x}}^{2} + 1$

Step 3.3

Rewrite ${\sqrt{x}}^{2}$ as $x$ .

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Step 3.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

$u_{lower} = {(x^{\frac{1}{2}})}^{2} + 1$

Step 3.3.2

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

$u_{lower} = x^{\frac{1}{2} \cdot 2} + 1$

Step 3.3.3

Combine $\frac{1}{2}$ and $2$ .

$u_{lower} = x^{\frac{2}{2}} + 1$

Step 3.3.4

Cancel the common factor of $2$ .

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Step 3.3.4.1

Cancel the common factor.

$u_{lower} = x^{\frac{2}{2}} + 1$

Step 3.3.4.2

Rewrite the expression.

$u_{lower} = x^{1} + 1$

Step 3.3.5

Simplify.

$u_{lower} = x + 1$

Step 3.4

Substitute the upper limit in for $t$ in $u = t^{2} + 1$ .

$u_{upper} = {(2 x)}^{2} + 1$

Step 3.5

Simplify each term.

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Step 3.5.1

Apply the product rule to $2 x$ .

$u_{upper} = 2^{2} x^{2} + 1$

Step 3.5.2

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

$u_{upper} = 4 x^{2} + 1$

Step 3.6

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

$u_{lower} = x + 1$

$u_{upper} = 4 x^{2} + 1$

Step 3.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$\arctan (t) t]_{\sqrt{x}}^{2 x} - \int_{x + 1}^{4 x^{2} + 1} \frac{1}{u} \cdot \frac{1}{2} d u$

Step 4

Simplify.

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Step 4.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$\arctan (t) t]_{\sqrt{x}}^{2 x} - \int_{x + 1}^{4 x^{2} + 1} \frac{1}{u \cdot 2} d u$

Step 4.2

Move $2$ to the left of $u$ .

$\arctan (t) t]_{\sqrt{x}}^{2 x} - \int_{x + 1}^{4 x^{2} + 1} \frac{1}{2 u} d u$

Step 5

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$\arctan (t) t]_{\sqrt{x}}^{2 x} - (\frac{1}{2} \int_{x + 1}^{4 x^{2} + 1} \frac{1}{u} d u)$

Step 6

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

$\arctan (t) t]_{\sqrt{x}}^{2 x} - \frac{1}{2} \ln (| u |)]_{x + 1}^{4 x^{2} + 1}$

Step 7

Substitute and simplify.

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Step 7.1

Evaluate $\arctan (t) t$ at $2 x$ and at $\sqrt{x}$ .

$(\arctan (2 x) (2 x)) - \arctan (\sqrt{x}) \sqrt{x} - \frac{1}{2} \ln (| u |)]_{x + 1}^{4 x^{2} + 1}$

Step 7.2

Evaluate $\ln (| u |)$ at $4 x^{2} + 1$ and at $x + 1$ .

$(\arctan (2 x) (2 x)) - \arctan (\sqrt{x}) \sqrt{x} - \frac{1}{2} ((\ln (| 4 x^{2} + 1 |)) - \ln (| x + 1 |))$

Step 7.3

Remove parentheses.

$\arctan (2 x) (2 x) - \arctan (\sqrt{x}) \sqrt{x} - \frac{1}{2} (\ln (| 4 x^{2} + 1 |) - \ln (| x + 1 |))$

Step 8

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

$\arctan (2 x) (2 x) - \arctan (\sqrt{x}) \sqrt{x} - \frac{1}{2} \ln (\frac{| 4 x^{2} + 1 |}{| x + 1 |})$

Step 9

Simplify.

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Step 9.1

Simplify each term.

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Step 9.1.1

Rewrite using the commutative property of multiplication.

$2 \arctan (2 x) x - \arctan (\sqrt{x}) \sqrt{x} - \frac{1}{2} \ln (\frac{| 4 x^{2} + 1 |}{| x + 1 |})$

Step 9.1.2

Combine $\ln (\frac{| 4 x^{2} + 1 |}{| x + 1 |})$ and $\frac{1}{2}$ .

$2 \arctan (2 x) x - \arctan (\sqrt{x}) \sqrt{x} - \frac{\ln (\frac{| 4 x^{2} + 1 |}{| x + 1 |})}{2}$

Step 9.2

To write $2 \arctan (2 x) x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- \arctan (\sqrt{x}) \sqrt{x} + 2 \arctan (2 x) x \cdot \frac{2}{2} - \frac{\ln (\frac{| 4 x^{2} + 1 |}{| x + 1 |})}{2}$

Step 9.3

Combine $2 \arctan (2 x) x$ and $\frac{2}{2}$ .

$- \arctan (\sqrt{x}) \sqrt{x} + \frac{2 \arctan (2 x) x \cdot 2}{2} - \frac{\ln (\frac{| 4 x^{2} + 1 |}{| x + 1 |})}{2}$

Step 9.4

Combine the numerators over the common denominator.

$- \arctan (\sqrt{x}) \sqrt{x} + \frac{2 \arctan (2 x) x \cdot 2 - \ln (\frac{| 4 x^{2} + 1 |}{| x + 1 |})}{2}$

Step 9.5

Multiply $2$ by $2$ .

$- \arctan (\sqrt{x}) \sqrt{x} + \frac{4 \arctan (2 x) x - \ln (\frac{| 4 x^{2} + 1 |}{| x + 1 |})}{2}$