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

Jan 8, 2017

The answer is $= {x}^{3} / 3 - x + 2 \arctan x + C$

#### Explanation:

Since the degree of the numerator is not less than the degree of the denominator, perform a 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}$$+ 1$$\textcolor{w h i t e}{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 a a a}$color(white)(aa)∣${x}^{2} - 1$

$\textcolor{w h i t e}{a a a a}$$0 - {x}^{2}$$\textcolor{w h i t e}{a a a a}$$+ 1$

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

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

Therefore,

$\frac{{x}^{4} + 1}{{x}^{2} + 1} = {x}^{2} - 1 + \frac{2}{{x}^{2} + 1}$

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

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

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

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

Therefore,

$2 \int \frac{\mathrm{dx}}{{x}^{2} + 1} = 2 \int \frac{{\sec}^{2} \theta d \theta}{\sec} ^ 2 \theta = 2 \int d \theta = 2 \theta = 2 \arctan x$

So,

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