Evaluate integral of 1/(x^2-9) with respect to x

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

Factor the numerator and denominator of $\frac{1}{x^{2} - 9}$ .

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Rewrite $9$ as $3^{2}$ .

$\int \frac{1}{x^{2} - 3^{2}} d x$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$\int \frac{1}{(x + 3) (x - 3)} d x$

Write the fraction using partial fraction decomposition.

$\int \frac{A_{1}}{x + 3} + \frac{A_{2}}{x - 3} d x$

Simplify.

$\int - \frac{1}{6 (x + 3)} + \frac{1}{6 (x - 3)} d x$

Split the single integral into multiple integrals.

$\int - \frac{1}{6 (x + 3)} d x + \int \frac{1}{6 (x - 3)} d x$

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$- \int \frac{1}{6 (x + 3)} d x + \int \frac{1}{6 (x - 3)} d x$

Since $\frac{1}{6}$ is constant with respect to $x$ , move $\frac{1}{6}$ out of the integral.

$- (\frac{1}{6} \int \frac{1}{x + 3} d x) + \int \frac{1}{6 (x - 3)} d x$

Let $u_{1} = x + 3$ . Then $d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Let $u_{1} = x + 3$ . Find $\frac{d u_{1}}{d x}$ .

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Differentiate $x + 3$ .

$\frac{d}{d x} [x + 3]$

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [3]$

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

$1 + \frac{d}{d x} [3]$

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$1 + 0$

Add $1$ and $0$ .

$1$

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

$- \frac{1}{6} \int \frac{1}{u_{1}} d u_{1} + \int \frac{1}{6 (x - 3)} d x$

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

$- \frac{1}{6} (\ln (| u_{1} |) + C) + \int \frac{1}{6 (x - 3)} d x$

Since $\frac{1}{6}$ is constant with respect to $x$ , move $\frac{1}{6}$ out of the integral.

$- \frac{1}{6} (\ln (| u_{1} |) + C) + \frac{1}{6} \int \frac{1}{x - 3} d x$

Let $u_{2} = x - 3$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Let $u_{2} = x - 3$ . Find $\frac{d u_{2}}{d x}$ .

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Differentiate $x - 3$ .

$\frac{d}{d x} [x - 3]$

By the Sum Rule, the derivative of $x - 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$

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

$1 + \frac{d}{d x} [- 3]$

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$1 + 0$

Add $1$ and $0$ .

$1$

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$- \frac{1}{6} (\ln (| u_{1} |) + C) + \frac{1}{6} \int \frac{1}{u_{2}} d u_{2}$

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

$- \frac{1}{6} (\ln (| u_{1} |) + C) + \frac{1}{6} (\ln (| u_{2} |) + C)$

Simplify.

$- \frac{1}{6} \ln (| u_{1} |) + \frac{1}{6} \ln (| u_{2} |) + C$

Substitute back in for each integration substitution variable.

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Replace all occurrences of $u_{1}$ with $x + 3$ .

$- \frac{1}{6} \ln (| x + 3 |) + \frac{1}{6} \ln (| u_{2} |) + C$

Replace all occurrences of $u_{2}$ with $x - 3$ .

$- \frac{1}{6} \ln (| x + 3 |) + \frac{1}{6} \ln (| x - 3 |) + C$

Evaluate integral of 1/(x^2-9) with respect to x

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