Question

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

• A. $\frac{1}{2}\log \left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+c$
• B. $\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|-\frac{4}{x-1}+c$
  
• C. $\frac{1}{2}\log \left|\frac{x+1}{x-1}\right|+\frac{4}{x-1}+c$
  
• D. None of these

Answer 1

The correct option is A $\frac{1}{2}\log \left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+c$

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

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

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

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

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

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

$3x+5=A\left(x^2-1\right)+Bx+B+C\left(x^2+1-2x\right)$

$\Rightarrow 3x+5=A{x}^2-A+Bx+B+c{x}^2+c-2cx$

$\Rightarrow 3x+5=\left(A+c\right)x^2+\left(B-2c\right)x-A+B+c$

On Comparing both sides we get

$A+c=0$ $B-2c=3$ $-A+B+c=5$

$\Rightarrow A=-c$ $\Rightarrow B=3+2c$

$-A+B+c=5$

$\Rightarrow c+3+2c+c=5$

$\Rightarrow 4c=2$

$\Rightarrow c=\frac{1}{2}$

$B=3+2c=3+2\times \frac{1}{2}=4$

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

$\int \frac{3x+5}{x^3-x^2-x+1}=\frac{-1}{2}\int \frac{dx}{dt}+4\int \frac{dx}{{\left(x-1\right)}^2}+\frac{1}{2}\int \frac{dx}{x+1}$

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

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