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

Sep 8, 2016

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

#### Explanation:

As in the given algebraic fraction, we have the only denominator $1 + {x}^{2}$, which cannot be factorized. As it is a quadratic function, in partial decomposition, the numerator would be in form $a x + b$. But it is already in this form. Hence we cannot convert them into further partial fractions.

Now let $u = 1 + {x}^{2}$, hence $\mathrm{du} = 2 x \mathrm{dx}$

Hence, $\int \frac{x - 1}{1 + {x}^{2}} \mathrm{dx}$

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

Now $\int \frac{x}{1 + {x}^{2}} \mathrm{dx} = \int \frac{\mathrm{du}}{2 u} = \frac{1}{2} \times \ln u = \frac{\ln \left(1 + {x}^{2}\right)}{2}$

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

(as if ${\tan}^{- 1} x = v$, $x = \tan v$ and

$\frac{\mathrm{dx}}{\mathrm{dv}} = {\sec}^{2} v = 1 + {\tan}^{2} v = 1 + {x}^{2}$, hence $\frac{\mathrm{dx}}{1 + {x}^{2}} = \mathrm{dv}$

and $\int \frac{\mathrm{dx}}{1 + {x}^{2}} = \int \mathrm{dv} = v = {\tan}^{- 1} x$)

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