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

$\int \frac{1}{x^{3} + x} d x$

Factor $x$ out of $x^{3} + x$ .

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Factor $x$ out of $x^{3}$ .

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

Raise $x$ to the power of $1$ .

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

Factor $x$ out of $x^{1}$ .

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

Factor $x$ out of $x \cdot x^{2} + x \cdot 1$ .

$\int \frac{1}{x (x^{2} + 1)} d x$

Write the fraction using partial fraction decomposition.

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

Simplify.

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

Split the single integral into multiple integrals.

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

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

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

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

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

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

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

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Differentiate $x^{2} + 1$ .

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

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

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

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

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

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

$2 x + 0$

Add $2 x$ and $0$ .

$2 x$

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

$\ln (| x |) + C - \int \frac{1}{u} \cdot \frac{1}{2} d u$

Simplify.

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

$\ln (| x |) + C - \int \frac{1}{u \cdot 2} d u$

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

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

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

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

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

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

Simplify.

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

Replace all occurrences of $u$ with $x^{2} + 1$ .

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

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

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