# How do you integrate (x^3)/(1+x^2)?

Jun 15, 2016

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

#### Explanation:

First note that the given integrand (function to be integrated) is an Improper Rational Function [degree of poly. in Numerator (Nr.) $= 3 > 2 =$degree of poly. in Deno.(Dr.)]. So, before integrating, we have to make it Proper. Usually, Long Division is performed for this, but we proceed as under:

$N r . = {x}^{3} = {x}^{3} + x - x = x \left({x}^{2} + 1\right) - x .$
$\therefore$ Integrand = Nr./Dr. = $\frac{x \left({x}^{2} + 1\right) - x}{{x}^{2} + 1} = \frac{x \left({x}^{2} + 1\right)}{{x}^{2} + 1} - \frac{x}{{x}^{2} + 1} = x - \frac{x}{{x}^{2} + 1}$

$\therefore \int {x}^{3} / \left({x}^{2} + 1\right) \mathrm{dx} = \int \left[x - \frac{x}{{x}^{2} + 1}\right] \mathrm{dx} = \int x \mathrm{dx} - \left(\frac{1}{2}\right) \int \frac{2 x}{{x}^{2} + 1} \mathrm{dx} = {x}^{2} / 2 - \frac{1}{2} \ln \left({x}^{2} + 1\right) + C .$

Notice that the second integral follows by using the formula ; $\int \frac{f ' \left(x\right)}{f} \left(x\right) \mathrm{dx} = \ln f \left(x\right) .$

Jun 15, 2016

$\int {x}^{3} / \left(1 + {x}^{2}\right) d x = \frac{1}{2} \left[{x}^{2} - \ln \left(1 + {x}^{2}\right)\right] + C$

#### Explanation:

int x^3/(1+x^2) d x=?

(x^³)/(1+x^2)=x-x/(1+x^2)

$\text{we can write the expression of "x^3/(1+x^2) " as "x-x/(1+x^2)" because both expressions are equal}$

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

$\int {x}^{3} / \left(1 + {x}^{2}\right) d x = \frac{1}{2} {x}^{2} - \frac{1}{2} \int \frac{2 x \cdot d x}{1 + {x}^{2}}$

$1 + {x}^{2} = u \text{ ; } 2 x \cdot d x = d u$

$\int {x}^{3} / \left(1 + {x}^{2}\right) d x = \frac{1}{2} {x}^{2} - \frac{1}{2} \int \frac{d u}{u}$

$\int {x}^{3} / \left(1 + {x}^{2}\right) d x = \frac{1}{2} {x}^{2} - \frac{1}{2} \ln u + C$

"plug "u=1+x^2;

$\int {x}^{3} / \left(1 + {x}^{2}\right) d x = \frac{1}{2} {x}^{2} - \frac{1}{2} \ln \left(1 + {x}^{2}\right) + C$

$\int {x}^{3} / \left(1 + {x}^{2}\right) d x = \frac{1}{2} \left[{x}^{2} - \ln \left(1 + {x}^{2}\right)\right] + C$