# How do you find int (2x)/((1-x)(1+x^2)) dx using partial fractions?

Jan 9, 2017

$\int \frac{2 x \mathrm{dx}}{\left(1 - x\right) \left(1 + {x}^{2}\right)} = - \ln \left\mid 1 - x \right\mid + \frac{1}{2} \ln \left(1 + {x}^{2}\right) - \arctan x$

#### Explanation:

Develop the integral in partial fractions:

$\frac{2 x}{\left(1 - x\right) \left(1 + {x}^{2}\right)} = \frac{A}{1 - x} + \frac{B x + C}{1 + {x}^{2}}$

$\frac{2 x}{\left(1 - x\right) \left(1 + {x}^{2}\right)} = \frac{A \left(1 + {x}^{2}\right) + \left(B x + C\right) \left(1 - x\right)}{\left(1 - x\right) \left(1 + {x}^{2}\right)}$

$2 x = A + A {x}^{2} + B x + C - B {x}^{2} - C x$

$2 x = \left(A - B\right) {x}^{2} + \left(B - C\right) x + \left(A + C\right)$

So:

$A - B = 0 \implies A = B$

$B - C = 2 \implies A - C = 2$

$\left[\begin{matrix}A - C = 2 \\ A + C = 0\end{matrix}\right] \implies A = 1 , B = 1 , C = - 1$

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

$\int \frac{2 x \mathrm{dx}}{\left(1 - x\right) \left(1 + {x}^{2}\right)} = \int \frac{\mathrm{dx}}{1 - x} + \int \frac{x \mathrm{dx}}{1 + {x}^{2}} - \int \frac{\mathrm{dx}}{1 + {x}^{2}} = - \ln \left\mid 1 - x \right\mid + \frac{1}{2} \ln \left(1 + {x}^{2}\right) - \arctan x$