Question
Evaluate :
$
\int \frac{x^3}{(x-1)\left(x^2+1\right)} d x
$

Answer

$
\begin{array}{l}
\int \frac{x^3}{(x-1)\left(x^2-1\right)} d x \\
\Rightarrow \int \frac{x^3}{x^3-x^2+x-1} d x
\end{array}
$
Here degree of numerator and denominator is same. So on dividing numerator by denominator :
$
\begin{array}{l}
\int 1+\frac{x^2-x+1}{(x-1)\left(x^2+1\right)} d x \\
\int d x+\int \frac{x^2-x+1}{(x-1)\left(x^2+1\right)} d x
\end{array}
$
Using Partial fraction,
$
\begin{array}{l}
\frac{x^2-x+1}{(x-1)\left(x^2+1\right)}=\frac{A}{(x-1)}+\frac{B x+C}{x^2+1} \\
\Rightarrow x^2-x+1=A\left(x^2+1\right)+(B x+C)(x-1)
\end{array}
$
Putting $x=1 \Rightarrow 1=2 A \quad \therefore A =\frac{1}{2}$
Comparing coefficient of $x^2 \Rightarrow 1= A + B \therefore B =1-\frac{1}{2}$ $=\frac{1}{2}$
Comparing coefficient of $x-1=- B + C \therefore C =-\frac{1}{2}$
$
1+\frac{1}{2(x-1)}+\frac{\frac{1}{2} x-\frac{1}{2}}{x^2+1}
$
$
1+\frac{1}{2(x-1)}+\frac{1}{2} \frac{x}{x^2+1}-\frac{1}{2} \cdot \frac{1}{x^2+1}
$
On integrating,
$
\begin{array}{l}
\int 1 d x+\frac{1}{2} \int \frac{1}{(x-1)}+\frac{1}{2} \int \frac{x}{x^2+1} d x-\frac{1}{2} \int \frac{1}{x^2+1} d x \\
=x+\frac{1}{2} \log |x-1|+\frac{1}{4} \log \left|x^2+1\right|-\frac{1}{2} \tan ^{-1} x+C
\end{array}
$

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