# What is the antiderivative of x^3/(1+x^2)?

Jul 7, 2016

$\frac{1}{2} \left\{{x}^{2} + 1 - \ln \left({x}^{2} + 1\right)\right\} + C .$

OR

$\frac{1}{2} \left\{{x}^{2} - \ln \left({x}^{2} + 1\right)\right\} + {C}_{1} ,$ where ${C}_{1} = C + \frac{1}{2.}$

#### Explanation:

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

We take substn. ${x}^{2} + 1 = t$, so that, $2 x \mathrm{dx} = \mathrm{dt} .$

Also, ${x}^{2} + 1 = t \Rightarrow {x}^{2} = t - 1$

Now, $I = \int {x}^{3} / \left(1 + {x}^{2}\right) \mathrm{dx} = \frac{1}{2} \int \frac{{x}^{2} \cdot 2 x}{1 + {x}^{2}} \mathrm{dx} = \frac{1}{2} \int \frac{t - 1}{t} \mathrm{dt}$
$= \frac{1}{2} \int \left\{\frac{t}{t} - \frac{1}{t}\right\} \mathrm{dt} = \frac{1}{2} \int \left\{1 - \frac{1}{t}\right\} \mathrm{dt} = \frac{1}{2} \left\{t - \ln t\right\} = \frac{1}{2} \left\{{x}^{2} + 1 - \ln \left({x}^{2} + 1\right)\right\} + C .$

$I$ is also$= \frac{1}{2} {x}^{2} + \frac{1}{2} - \frac{1}{2} \ln \left({x}^{2} + 1\right) + C$
$= \frac{1}{2} \left\{{x}^{2} - \ln \left({x}^{2} + 1\right)\right\} + {C}_{1} ,$ where ${C}_{1} = C + \frac{1}{2}$

Is this not enjoyable?! Enjoy maths.!