# How do I find the anti derivative of (2x^2+5)/(x^2+1)?

Nov 27, 2016

Rewrite the integrand.

#### Explanation:

$\frac{2 {x}^{2} + 5}{{x}^{2} + 1} = \frac{2 {x}^{2} + 2 + 3}{{x}^{2} + 1}$

$= \frac{2 {x}^{2} + 2}{{x}^{2} + 1} + \frac{3}{{x}^{2} + 1}$

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

So the antiderivative is

$2 x + 3 {\tan}^{-} 1 x + C$

If you haven't memorized $\int \frac{1}{{x}^{2} + 1} \mathrm{dx} = {\tan}^{-} 1 x + C$, then you can use trigonometric substitution

$x = \tan \theta$ so $\mathrm{dx} = {\sec}^{2} \theta d \theta$ (and $x = {\tan}^{-} 1 \theta$)

After substitution, the integral becomes

$\int \frac{1}{{\tan}^{2} \theta + 1} {\sec}^{2} \theta d \theta = \int \frac{1}{\sec} ^ 2 \theta {\sec}^{2} \theta d \theta$

$= \int 1 d \theta = \theta + C = {\tan}^{-} 1 x + C$

If you don't notice how to split the numerator then do the division.

$\left(2 {x}^{2} + 5\right) \div \left({x}^{2} + 1\right) = 2 + \frac{3}{{x}^{2} + 1}$

When finding the antiderivative (integral) of a rational function, if the degree of the numerator is greater than the degree of its denominator, division is often a good idea.