Question

Integrate the rational function:
$\frac{3x+5}{x^3-x^2-x+1}$

Answer 1

$\int \frac{3x+5}{x^3-x^2-x+1}.dx$

We know,

$\frac{3x+5}{x^3-x^2-x+1}$

$=\frac{3x+5}{(x-1)(x^2-1)}$

$=\frac{3x+5}{(x-1)(x-1)(x+1)}$

$=\frac{3x+5}{(x+1)(x-1{)}^2}$

Using partial fraction

$\frac{(3x+5)}{(x+1)(x-1{)}^2}$

$=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1{)}^2}$

$\Rightarrow 3x+5=A(x-1{)}^2+B(x^2-1)+C(x+1)$

Put $x=1$

$\Rightarrow 3\times 1+5$

$=A(1-1{)}^2+B(1^2-1)+C(1+1)$

$\Rightarrow 8=2C$

$\Rightarrow C=4$

Putting $x=-1$

$\Rightarrow 3(-1)+5$

$=A(-1-1{)}^2+B((-1{)}^2-1)+C(-1+1)$

$\Rightarrow -3+5=A(-2{)}^2$

Putting $x=0$

$\Rightarrow 3\left(0\right)+5$

$=A(0-1{)}^2+B(0-1)+C(0-1)$

$\Rightarrow 5=A-B+C$

$\Rightarrow 5=\frac{1}{2}-B+4$

$[\because A=\frac{1}{2},C=4]$

$\Rightarrow 1-\frac{1}{2}=-B\Rightarrow B=-\frac{1}{2}$

Putting the values of $A,B$ and $C$

$\frac{3x+5}{(x+1)(x-1{)}^2}=\frac{1}{2(x+1)}-\frac{1}{2(x-1)}+\frac{4}{(x-1{)}^2}$

Now,
$\int \frac{3x+5}{x^3-x^2-x+1}.dx=\int [\frac{1}{2(x+1)}-\frac{1}{2(x-1)}+\frac{4}{(x-1{)}^2}].dx$

$=\frac{1}{2}\int \frac{dx}{x+1}-\frac{1}{2}\int \frac{dx}{x-1}+4\int \frac{dx}{(x-1{)}^2}$

$=\frac{1}{2}\log |x+1|-\frac{1}{2}\log |x-1|+4\int (x-1{)}^{-2}dx+C$

$=\frac{1}{2}\log \left|\frac{x+1}{x-1}\right|+4\left[\frac{(x-1{)}^{-2+1}}{-2+1}\right]+C$

$=\frac{1}{2}\log \left|\frac{x+1}{x-1}\right|-\frac{4}{(x-1)}+C$

Where $C$ is constant of integration