# How do you find the indefinite integral of int x^2/(3-x^2)?

Jan 23, 2017

$= - x + \sqrt{3} {\tanh}^{- 1} \left(\frac{x}{\sqrt{3}}\right) + C$

#### Explanation:

$\int {x}^{2} / \left(3 - {x}^{2}\right) \textcolor{red}{\mathrm{dx}}$

$= \int \frac{\left(- 3 + {x}^{2}\right) + 3}{3 - {x}^{2}} \mathrm{dx}$

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

$= - x + \textcolor{b l u e}{\int \frac{3}{3 - {x}^{2}} \mathrm{dx}} \star$

For the blue bit, we will use the hyperbolic identity:

$1 - {\tanh}^{2} y = {\sech}^{2} y$

So we let $x = \sqrt{3} \tanh y \implies \mathrm{dx} = \sqrt{3} {\sech}^{2} y \mathrm{dy}$ so the blue part of $\star$ becomes:

$\int \frac{3}{3 - 3 {\tanh}^{2} y} \sqrt{3} {\sech}^{2} y \setminus \mathrm{dy}$

$= \int \frac{3}{3 - 3 {\tanh}^{2} y} \sqrt{3} {\sech}^{2} y \setminus \mathrm{dy}$

$= \int \frac{1}{\sech} ^ 2 y \sqrt{3} {\sech}^{2} y \setminus \mathrm{dy}$

$= \sqrt{3} \int \mathrm{dy}$

$= \sqrt{3} {\tanh}^{- 1} \left(\frac{x}{\sqrt{3}}\right) + C$

Feeding this into $\star$

$\implies - x + \sqrt{3} {\tanh}^{- 1} \left(\frac{x}{\sqrt{3}}\right) + C$