# How do you integrate (x-1)/ (x^3 +x^2) using partial fractions?

Aug 17, 2016

We find that, the $D r . = {x}^{3} + x = x \left({x}^{2} + 1\right)$

This, the poly. of the $D r .$ has a linear factor $x$, and a non-reducible quadr. factor ${x}^{2} + 1$. Therefore, we will suppose that

$\frac{x - 1}{{x}^{3} + x} = \frac{x - 1}{x \left({x}^{2} + 1\right)} = \frac{A}{x} + \frac{B x + c}{{x}^{2} + 1} , A , B , C \in \mathbb{R} \ldots \left(\star\right)$.

But, upon simplification,

$T h e R . H . S . = \frac{A \left({x}^{2} + 1\right) + \left(B x + C\right) x}{x \left({x}^{2} + 1\right)}$.

Thus, $\frac{x - 1}{{x}^{3} + x} = \frac{A \left({x}^{2} + 1\right) + \left(B x + C\right) x}{x \left({x}^{2} + 1\right)} \ldots \ldots \ldots \left(1\right)$.

As the $D r s .$ of $\left(1\right)$ are same, so must be the $N r s .$. Hence,

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

Comparing the resp. co-effs. of both sides, we immediately get,

$A = - 1 , C = 1 , \mathmr{and} , A + B = 0 \Rightarrow B = - A = 1$.

Therefore, by $\left(\star\right)$,

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

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

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

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

$= - \ln | x | + \frac{1}{2} \ln \left({x}^{2} + 1\right) + \arctan x + C$, OR,

$= \ln \left(\frac{\sqrt{{x}^{2} + 1}}{|} x |\right) + \arctan x + C$.

In the final step, this well-known Rule has been used :-

The Rule$: = \int \frac{f ' \left(x\right)}{f} \left(x\right) \mathrm{dx} = \ln | f \left(x\right) + K$.

I hope, this will be of Help! Enjoy Maths.!