Evaluate integral of 1/((x-1)(x+3)) with respect to x

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

Reorder the factors of $(x - 1) (x + 3)$ .

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

Write the fraction using partial fraction decomposition.

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

Simplify.

$\int \frac{1}{4 (x - 1)} - \frac{1}{4 (x + 3)} d x$

Split the single integral into multiple integrals.

$\int \frac{1}{4 (x - 1)} d x + \int - \frac{1}{4 (x + 3)} d x$

Since $\frac{1}{4}$ is constant with respect to $x$ , move $\frac{1}{4}$ out of the integral.

$\frac{1}{4} \int \frac{1}{x - 1} d x + \int - \frac{1}{4 (x + 3)} d x$

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

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

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Differentiate $x - 1$ .

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

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

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

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

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

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

$1 + 0$

Add $1$ and $0$ .

$1$

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

$\frac{1}{4} \int \frac{1}{u_{1}} d u_{1} + \int - \frac{1}{4 (x + 3)} d x$

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

$\frac{1}{4} (\ln (| u_{1} |) + C) + \int - \frac{1}{4 (x + 3)} d x$

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

$\frac{1}{4} (\ln (| u_{1} |) + C) - \int \frac{1}{4 (x + 3)} d x$

Since $\frac{1}{4}$ is constant with respect to $x$ , move $\frac{1}{4}$ out of the integral.

$\frac{1}{4} (\ln (| u_{1} |) + C) - (\frac{1}{4} \int \frac{1}{x + 3} d x)$

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 $d u_{2}$ .

$\frac{1}{4} (\ln (| u_{1} |) + C) - \frac{1}{4} \int \frac{1}{u_{2}} d u_{2}$

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

$\frac{1}{4} (\ln (| u_{1} |) + C) - \frac{1}{4} (\ln (| u_{2} |) + C)$

Simplify.

$\frac{1}{4} \ln (| u_{1} |) - \frac{1}{4} \ln (| u_{2} |) + C$

Substitute back in for each integration substitution variable.

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

$\frac{1}{4} \ln (| x - 1 |) - \frac{1}{4} \ln (| u_{2} |) + C$

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

$\frac{1}{4} \ln (| x - 1 |) - \frac{1}{4} \ln (| x + 3 |) + C$

Evaluate integral of 1/((x-1)(x+3)) with respect to x

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