Calculus Examples

$\int \frac{2 x + 1}{9 + 16 x^{2}} d x$

Step 1

Split the fraction $\frac{2 x + 1}{9 + 16 x^{2}}$ into two fractions.

$\int \frac{2 x}{9 + 16 x^{2}} + \frac{1}{9 + 16 x^{2}} d x$

Step 2

Split the single integral into multiple integrals.

$\int \frac{2 x}{9 + 16 x^{2}} d x + \int \frac{1}{9 + 16 x^{2}} d x$

Step 3

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

$2 \int \frac{x}{9 + 16 x^{2}} d x + \int \frac{1}{9 + 16 x^{2}} d x$

Step 4

Let $u = 9 + 16 x^{2}$ . Then $d u = 32 x d x$ , so $\frac{1}{32} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.1

Let $u = 9 + 16 x^{2}$ . Find $\frac{d u}{d x}$ .

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Step 4.1.1

Differentiate $9 + 16 x^{2}$ .

$\frac{d}{d x} [9 + 16 x^{2}]$

Step 4.1.2

Differentiate.

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Step 4.1.2.1

By the Sum Rule, the derivative of $9 + 16 x^{2}$ with respect to $x$ is $\frac{d}{d x} [9] + \frac{d}{d x} [16 x^{2}]$ .

$\frac{d}{d x} [9] + \frac{d}{d x} [16 x^{2}]$

Step 4.1.2.2

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [16 x^{2}]$

Step 4.1.3

Evaluate $\frac{d}{d x} [16 x^{2}]$ .

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Step 4.1.3.1

Since $16$ is constant with respect to $x$ , the derivative of $16 x^{2}$ with respect to $x$ is $16 \frac{d}{d x} [x^{2}]$ .

$0 + 16 \frac{d}{d x} [x^{2}]$

Step 4.1.3.2

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

$0 + 16 (2 x)$

Step 4.1.3.3

Multiply $2$ by $16$ .

$0 + 32 x$

Step 4.1.4

Add $0$ and $32 x$ .

$32 x$

Step 4.2

Rewrite the problem using $u$ and $d u$ .

$2 \int \frac{1}{u} \cdot \frac{1}{32} d u + \int \frac{1}{9 + 16 x^{2}} d x$

Step 5

Simplify.

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Step 5.1

Multiply $\frac{1}{u}$ by $\frac{1}{32}$ .

$2 \int \frac{1}{u \cdot 32} d u + \int \frac{1}{9 + 16 x^{2}} d x$

Step 5.2

Move $32$ to the left of $u$ .

$2 \int \frac{1}{32 u} d u + \int \frac{1}{9 + 16 x^{2}} d x$

Step 6

Since $\frac{1}{32}$ is constant with respect to $u$ , move $\frac{1}{32}$ out of the integral.

$2 (\frac{1}{32} \int \frac{1}{u} d u) + \int \frac{1}{9 + 16 x^{2}} d x$

Step 7

Simplify.

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Step 7.1

Combine $\frac{1}{32}$ and $2$ .

$\frac{2}{32} \int \frac{1}{u} d u + \int \frac{1}{9 + 16 x^{2}} d x$

Step 7.2

Cancel the common factor of $2$ and $32$ .

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Step 7.2.1

Factor $2$ out of $2$ .

$\frac{2 (1)}{32} \int \frac{1}{u} d u + \int \frac{1}{9 + 16 x^{2}} d x$

Step 7.2.2

Cancel the common factors.

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Step 7.2.2.1

Factor $2$ out of $32$ .

$\frac{2 \cdot 1}{2 \cdot 16} \int \frac{1}{u} d u + \int \frac{1}{9 + 16 x^{2}} d x$

Step 7.2.2.2

Cancel the common factor.

$\frac{2 \cdot 1}{2 \cdot 16} \int \frac{1}{u} d u + \int \frac{1}{9 + 16 x^{2}} d x$

Step 7.2.2.3

Rewrite the expression.

$\frac{1}{16} \int \frac{1}{u} d u + \int \frac{1}{9 + 16 x^{2}} d x$

Step 8

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

$\frac{1}{16} (\ln (| u |) + C) + \int \frac{1}{9 + 16 x^{2}} d x$

Step 9

Factor $16$ out of $9 + 16 x^{2}$ .

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Step 9.1

Factor $16$ out of $9$ .

$\frac{1}{16} (\ln (| u |) + C) + \int \frac{1}{16 (\frac{9}{16}) + 16 x^{2}} d x$

Step 9.2

Factor $16$ out of $16 x^{2}$ .

$\frac{1}{16} (\ln (| u |) + C) + \int \frac{1}{16 (\frac{9}{16}) + 16 (x^{2})} d x$

Step 9.3

Factor $16$ out of $16 (\frac{9}{16}) + 16 (x^{2})$ .

$\frac{1}{16} (\ln (| u |) + C) + \int \frac{1}{16 (\frac{9}{16} + x^{2})} d x$

Step 10

Since $\frac{1}{16}$ is constant with respect to $x$ , move $\frac{1}{16}$ out of the integral.

$\frac{1}{16} (\ln (| u |) + C) + \frac{1}{16} \int \frac{1}{\frac{9}{16} + x^{2}} d x$

Step 11

Rewrite $\frac{9}{16}$ as ${(\frac{3}{4})}^{2}$ .

$\frac{1}{16} (\ln (| u |) + C) + \frac{1}{16} \int \frac{1}{{(\frac{3}{4})}^{2} + x^{2}} d x$

Step 12

The integral of $\frac{1}{{(\frac{3}{4})}^{2} + x^{2}}$ with respect to $x$ is $\frac{1}{\frac{3}{4}} \arctan (\frac{x}{\frac{3}{4}}) + C$ .

$\frac{1}{16} (\ln (| u |) + C) + \frac{1}{16} (\frac{1}{\frac{3}{4}} \arctan (\frac{x}{\frac{3}{4}}) + C)$

Step 13

Simplify.

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Step 13.1

Simplify.

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Step 13.1.1

Multiply by the reciprocal of the fraction to divide by $\frac{3}{4}$ .

$\frac{1}{16} (\ln (| u |) + C) + \frac{1}{16} (1 (\frac{4}{3}) \arctan (\frac{x}{\frac{3}{4}}) + C)$

Step 13.1.2

Multiply $\frac{4}{3}$ by $1$ .

$\frac{1}{16} (\ln (| u |) + C) + \frac{1}{16} (\frac{4}{3} \arctan (\frac{x}{\frac{3}{4}}) + C)$

Step 13.1.3

Multiply by the reciprocal of the fraction to divide by $\frac{3}{4}$ .

$\frac{1}{16} (\ln (| u |) + C) + \frac{1}{16} (\frac{4}{3} \arctan (x \frac{4}{3}) + C)$

Step 13.1.4

Combine $x$ and $\frac{4}{3}$ .

$\frac{1}{16} (\ln (| u |) + C) + \frac{1}{16} (\frac{4}{3} \arctan (\frac{x \cdot 4}{3}) + C)$

Step 13.1.5

Move $4$ to the left of $x$ .

$\frac{1}{16} (\ln (| u |) + C) + \frac{1}{16} (\frac{4}{3} \arctan (\frac{4 \cdot x}{3}) + C)$

Step 13.1.6

Combine $\frac{4}{3}$ and $\arctan (\frac{4 x}{3})$ .

$\frac{1}{16} (\ln (| u |) + C) + \frac{1}{16} (\frac{4 \arctan (\frac{4 x}{3})}{3} + C)$

Step 13.2

Simplify.

$\frac{1}{16} \ln (| u |) + \frac{1}{16} \cdot \frac{4}{3} \arctan (\frac{4}{3} x) + C$

Step 13.3

Simplify.

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Step 13.3.1

Multiply $\frac{1}{16}$ by $\frac{4}{3}$ .

$\frac{1}{16} \ln (| u |) + \frac{4}{16 \cdot 3} \arctan (\frac{4}{3} x) + C$

Step 13.3.2

Multiply $16$ by $3$ .

$\frac{1}{16} \ln (| u |) + \frac{4}{48} \arctan (\frac{4}{3} x) + C$

Step 13.3.3

Cancel the common factor of $4$ and $48$ .

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Step 13.3.3.1

Factor $4$ out of $4$ .

$\frac{1}{16} \ln (| u |) + \frac{4 (1)}{48} \arctan (\frac{4}{3} x) + C$

Step 13.3.3.2

Cancel the common factors.

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Step 13.3.3.2.1

Factor $4$ out of $48$ .

$\frac{1}{16} \ln (| u |) + \frac{4 \cdot 1}{4 \cdot 12} \arctan (\frac{4}{3} x) + C$

Step 13.3.3.2.2

Cancel the common factor.

$\frac{1}{16} \ln (| u |) + \frac{4 \cdot 1}{4 \cdot 12} \arctan (\frac{4}{3} x) + C$

Step 13.3.3.2.3

Rewrite the expression.

$\frac{1}{16} \ln (| u |) + \frac{1}{12} \arctan (\frac{4}{3} x) + C$

Step 14

Replace all occurrences of $u$ with $9 + 16 x^{2}$ .

$\frac{1}{16} \ln (| 9 + 16 x^{2} |) + \frac{1}{12} \arctan (\frac{4}{3} x) + C$