Evaluate the following integral:

Question:

Evaluate the following integral:

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

Solution:

Denominator is factorized, so let separate the fraction through partial fraction, hence let

$\frac{\mathrm{x}}{\left(\mathrm{x}^{2}+1\right)(\mathrm{x}-1)}=\frac{\mathrm{Ax}+\mathrm{B}}{\left(\mathrm{x}^{2}+1\right)}+\frac{\mathrm{Cx}+\mathrm{D}}{\mathrm{x}-1} \ldots \ldots$ (i)

$\Rightarrow \frac{\mathrm{x}}{\left(\mathrm{x}^{2}+1\right)(\mathrm{x}-1)}=\frac{(\mathrm{Ax}+\mathrm{B})(\mathrm{x}-1)+(\mathrm{Cx}+\mathrm{D})\left(\mathrm{x}^{2}+1\right)}{\left(\mathrm{x}^{2}+1\right)(\mathrm{x}-1)}$

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

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

$\Rightarrow x=(C) x^{2}+(A+D) x^{2}+(B-A+C) x+(D-B) \ldots \ldots(i i)$

By equating similar terms, we get

$C=0$............(iii)

$A+D=0 \Rightarrow A=-D$  .....(iv)

$B-A+C=1$

$\Rightarrow B-(-D)+0=2($ from equation(iii) and (iv))

$\Rightarrow B=2-D \ldots \ldots \ldots .(v)$

$D-B=0 \Rightarrow D-(2-D)=0 \Rightarrow 2 D=2 \Rightarrow D=1$

So equation(iv) becomes $A=-1$

So equation (v) becomes, $B=2-1=1$

We put the values of A, B, C, and D values back into our partial fractions in equation (i) and replace this as the integrand. We get

$\int\left[\frac{x}{\left(x^{2}+1\right)(x-1)}\right] d x$

$\Rightarrow \int \frac{\mathrm{Ax}+\mathrm{B}}{\left(\mathrm{x}^{2}+1\right)}+\frac{\mathrm{Cx}+\mathrm{D}}{\mathrm{x}-1} \mathrm{dx}$

$\Rightarrow \int\left[\frac{(-1) \mathrm{x}+1}{\left(\mathrm{x}^{2}+1\right)}+\frac{(0) \mathrm{x}+1}{\mathrm{x}-1}\right] \mathrm{dx}$

Split up the integral,

$\Rightarrow \int \frac{1}{\left(x^{2}+1\right)} d x-\int \frac{x}{\left(x^{2}+1\right)} d x+\int\left[\frac{1}{x-1}\right] d x$

Let substitute

$u=x^{2}+1 \Rightarrow d u=2 x d x \Rightarrow x d x=\frac{1}{2} d u$

$\mathrm{v}=\mathrm{x}-1 \Rightarrow \mathrm{dv}=\mathrm{dx}$

so the above equation becomes,

$\Rightarrow \int \frac{1}{\left(x^{2}+1\right)} d x-\frac{1}{2} \int \frac{1}{(u)} d u+\int\left[\frac{1}{v}\right] d v$

On integrating we get

$\Rightarrow \tan ^{-1} \mathrm{x}-\frac{1}{2} \log |\mathrm{u}|+\log |\mathrm{v}|+\mathrm{C}$

(the standard integral of $\frac{1}{\mathrm{x}^{2}+1}=\tan ^{-1} \mathrm{x}$ )

Substituting back, we get

$\Rightarrow \tan ^{-1} x-\frac{1}{2} \log \left|x^{2}+1\right|+\log |x-1|+C$

Note: the absolute value signs account for the domain of the natural $\log$ function $(x>0)$.

Hence,

$\int \frac{x}{\left(x^{2}+1\right)(x-1)} d x=\tan ^{-1} x-\frac{1}{2} \log \left|x^{2}+1\right|+\log |x-1|+C$