# How do you evaluate the integral int dx/(x^3+x)?

Jan 8, 2017

$\int \frac{\mathrm{dx}}{{x}^{3} + x} = \frac{1}{2} \ln \left({x}^{2} / \left({x}^{2} + 1\right)\right) + C$

#### Explanation:

We can write the integral as:

$\int \frac{\mathrm{dx}}{{x}^{3} + x} = \int \frac{\mathrm{dx}}{x \left(1 + {x}^{2}\right)}$

now we can substitute:

$x = \tan t$

$\mathrm{dx} = \frac{\mathrm{dt}}{\cos} ^ 2 t$

so we have:

$\int \frac{\mathrm{dx}}{{x}^{3} + x} = \int \frac{\mathrm{dt}}{{\cos}^{2} t \tan t \left(1 + {\tan}^{2} t\right)}$

and using the trigonometric identity:

$1 + {\tan}^{2} t = 1 + {\sin}^{2} \frac{t}{\cos} ^ 2 t = \frac{{\cos}^{2} t + {\sin}^{2} t}{\cos} ^ 2 t = \frac{1}{\cos} ^ 2 t$

$\int \frac{\mathrm{dx}}{{x}^{3} + x} = \int \frac{\mathrm{dt}}{{\cos}^{2} t \tan t \frac{1}{\cos} ^ 2 t} = \int \frac{\mathrm{dt}}{\tan} t = \int \frac{\cos t \mathrm{dt}}{\sin} t$

We can see that: $\cos t \mathrm{dt} = d \left(\sin t\right)$, so:

$\int \frac{\mathrm{dx}}{{x}^{3} + x} = \int \frac{\mathrm{ds} \int}{\sin} t = \ln \left\mid \sin \right\mid t + C$

To substitute back $x$ we can note that:

$x = \tan t = \sin \frac{t}{\sqrt{1 - {\sin}^{2} t}}$

${x}^{2} = {\sin}^{2} \frac{t}{1 - {\sin}^{2} t}$

${x}^{2} \left(1 - {\sin}^{2} t\right) = {\sin}^{2} t$

${x}^{2} - {x}^{2} {\sin}^{2} t = {\sin}^{2} t$

${x}^{2} = {x}^{2} {\sin}^{2} t + {\sin}^{2} t$

${x}^{2} = \left({x}^{2} + 1\right) {\sin}^{2} t$

${x}^{2} / \left({x}^{2} + 1\right) = {\sin}^{2} t$

$\sqrt{{x}^{2} / \left({x}^{2} + 1\right)} = \left\mid \sin \right\mid t$

so that finally:

$\int \frac{\mathrm{dx}}{{x}^{3} + x} = \ln \sqrt{{x}^{2} / \left({x}^{2} + 1\right)} + C = \frac{1}{2} \ln \left({x}^{2} / \left({x}^{2} + 1\right)\right) + C$