Question

The value of the integral $\int \frac{3x+5}{x^3-x^2-x+1}dx$ is
(where $C$ is integration constant)

• A. $\frac{1}{4}\ln \left|\frac{x+2}{x-1}\right|-\frac{4}{(x-1)}+C$
• B. $\frac{1}{4}\ln \left|\frac{x+1}{x-1}\right|-\frac{2}{(x+1)}+C$
• C. $\frac{1}{2}\ln \left|\frac{x+1}{x-1}\right|-\frac{4}{(x-1)}+C$
• D. $\frac{1}{2}\ln \left|\frac{x+1}{x-1}\right|+\frac{2}{(x-1)}+C$

Answer 1

The correct option is C $\frac{1}{2}\ln \left|\frac{x+1}{x-1}\right|-\frac{4}{(x-1)}+C$

$\frac{3x+5}{x^3-x^2-x+1}=\frac{3x+5}{(x-1{)}^2(x+1)}$
Let
$\frac{3x+5}{(x-1{)}^2(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1{)}^2}+\frac{C}{(x+1)}\Rightarrow 3x+5=A(x-1)(x+1)+B(x+1)+C(x-1{)}^2\Rightarrow 3x+5=A(x^2-1)+B(x+1)+C(x^2+1-2x)\cdots (1)$
Equating the coefficients of $x^2,x$ and constant term, we obtain
$A+C=0B-2C=3-A+B+C=5$
On solving, we obtain
$B=4,A=-\frac{1}{2},C=\frac{1}{2}\frac{3x+5}{(x-1{)}^2(x+1)}=\frac{-1}{2(x-1)}+\frac{4}{(x-1{)}^2}+\frac{1}{2(x+1)}\therefore \int \frac{3x+5}{(x-1{)}^2(x+1)}dx=\frac{-1}{2}\int \frac{1}{x-1}dx+4\int \frac{1}{(x-1{)}^2}dx+\frac{1}{2}\int \frac{1}{(x+1)}dx=-\frac{1}{2}\ln |x-1|+4\left(\frac{-1}{x-1}\right)+\frac{1}{2}\ln |x+1|+C=\frac{1}{2}\ln \left|\frac{x+1}{x-1}\right|-\frac{4}{(x-1)}+C$