How do you evaluate the integral int x^4/(1+x^2)dx?

Jan 18, 2017

The answer is $= {x}^{3} / 3 - x + \arctan \left(x\right) + C$

Explanation:

We need

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(n \ne - 1\right)$

We perform a polynomial long division

$\textcolor{w h i t e}{a a a a}$${x}^{4}$$\textcolor{w h i t e}{a a a a a a a a a}$∣${x}^{2} + 1$

$\textcolor{w h i t e}{a a a a}$${x}^{4} + {x}^{2}$$\textcolor{w h i t e}{a a a a a}$∣${x}^{2} - 1$

$\textcolor{w h i t e}{a a a a a}$$0 - {x}^{2}$

$\textcolor{w h i t e}{a a a a a a a}$$- {x}^{2} - 1$

$\textcolor{w h i t e}{a a a a a a a a a}$$0 + 1$

Therefore,

${x}^{4} / \left({x}^{2} + 1\right) = \left({x}^{2} - 1\right) + \frac{1}{{x}^{2} + 1}$

So,

$\int \frac{{x}^{4} \mathrm{dx}}{{x}^{2} + 1} = \int \left({x}^{2} - 1\right) \mathrm{dx} + \textcolor{red}{\int \frac{\mathrm{dx}}{{x}^{2} + 1}}$

$= {x}^{3} / 3 - x + \textcolor{red}{\int \frac{\mathrm{dx}}{{x}^{2} + 1}}$

The integral in red is done by trigonometric substitution

Let $x = \tan \theta$, $\implies$, $\mathrm{dx} = {\sec}^{2} \theta d \theta$

and, ${x}^{2} + 1 = {\tan}^{2} \theta + 1 = {\sec}^{2} \theta$

Therefore,

$\textcolor{red}{\int \frac{\mathrm{dx}}{{x}^{2} + 1} = \int \frac{{\sec}^{2} \theta d \theta}{\sec} ^ 2 \theta = \int d \theta = \theta = \arctan \left(x\right)}$

Putting it all together

$\int \frac{{x}^{4} \mathrm{dx}}{{x}^{2} + 1} = {x}^{3} / 3 - x + \arctan \left(x\right) + C$