How do you find the antiderivative of (5x^2)/(x^2 + 1)?

Feb 20, 2015

I would start by factorizing out ${x}^{2}$ from the denominator:

$\int \frac{5 {x}^{2}}{{x}^{2} + 1} \mathrm{dx} = \int \frac{5 {x}^{2}}{{x}^{2} \left(1 + \frac{1}{x} ^ 2\right)} \mathrm{dx} = \int \frac{5}{1 + \frac{1}{x} ^ 2} \mathrm{dx} =$

now set: $\frac{1}{x} ^ 2 = {t}^{2}$
$x = \frac{1}{t}$ and $\mathrm{dx} = - \frac{1}{t} ^ 2 \mathrm{dt}$

The integral becomes:

$5 \int \frac{1}{1 + {t}^{2}} \cdot \left(- \frac{1}{t} ^ 2\right) \mathrm{dt} = - 5 \left[\int \frac{1}{{t}^{2} \left(1 + {t}^{2}\right)} \mathrm{dt}\right] =$

$= - 5 \left\{\int \left[\frac{1}{t} ^ 2 - \frac{1}{1 + {t}^{2}}\right] \mathrm{dt}\right\} =$

$= - 5 \left[{t}^{- 1} / - 1 - \arctan \left(t\right)\right] =$ going back to $x$

$5 x + 5 \arctan \left(\frac{1}{x}\right) + c$