Evaluate : x^3 (x-1) (x^2+1 ) d x - Vidyadip

Question

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

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Answer

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

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