Question

Integrate the rational functions.
$\int \frac{x}{(x^2+1)(x-1)}dx.$

Answer 1

$\int \frac{x}{(x^2+1)(x-1)}dx$
First, we reslove the given integrand into partial fractions.
Let $\frac{x}{(x^2+1)(x-1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}......\left(i\right)$
$\Rightarrow x=A(x^2+1)+(Bx+C)(x-1).........\left(ii\right)$
Substituting x=1 and 0 in Eq. (ii), we get
$1=A\left(2\right)\text{and}0=A-C\Rightarrow A=\frac{1}{2}\text{and}C=A=\frac{1}{2}$
On equating the coefficient of $x^2$ on the both sides in Eq. (ii), we get
$0=A+B\Rightarrow B=-A=-\frac{1}{2}\therefore \int \frac{x}{(x^2+1)(x-1)}dx=\int \{\frac{\frac{1}{2}}{x-1}+\frac{(-\frac{1}{2}x+\frac{1}{2})}{x^2+1}\}dx=\frac{1}{2}\int \frac{1}{x-1}dx+\frac{1}{2}\int \frac{(-x+1)}{x^2+1}dx=\frac{1}{2}\int \frac{1}{x-1}dx-\frac{1}{2}\int \frac{x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx=\frac{1}{2}log|x-1|-\frac{1}{4}{I}_1+\frac{1}{2}ta{n}^{-1}x+C_1.......\left(iii\right)$
where, $I_1=\int \frac{2x}{x^2+1}dx.\text{Let}(x^2+1)=t\Rightarrow 2xdx=dt$

$\therefore I_1=\int \frac{dt}{t}=log|t|+C_2=log|x^2+1|+C_2$
On putting these values in Eq. (iii), we get
$\int \frac{x}{(x^2+1)(x-1)}dx=\frac{1}{2}log|x-1|-\frac{1}{4}log|x^2+1|+\frac{1}{2}ta{n}^{-1}x+C[C=C_1+C_2]$