Question

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

Answer 1

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

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

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

Solving by the method of partial fractions,

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

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

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

Put $x=1$

$\Rightarrow 8=2B$ or $B=4$

Put $x=-1$

$\Rightarrow -3+5=4C$ or $C=\frac{1}{2}$

Put $x=0$

$\Rightarrow 5=-A+B+C$

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

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

$\therefore A=-\frac{1}{2}$

Now,$\int \frac{(3x+5)dx}{x^3-x^2-x+1}$

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

$=-\frac{1}{2}\log |x-1|-\frac{4}{x-1}+\frac{1}{2}\log |x+1|+c$

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