Evaluate the following intregals: 3 (1-x)(1+x^2) dx

Question

Evaluate the following intregals:
$\int\frac{3}{(1-\text{x})(1+\text{x}^2)}\ \text{dx}$

✓

Answer

We have,
$\text{I}=\int\frac{3\text{dx}}{(1-\text{x})(1+\text{x}^2)}$
$=3\int\frac{\text{dx}}{(1-\text{x})(1+\text{x}^2)}$
Let $\frac{1}{(1-\text{x})(1+\text{x})^2}=\frac{\text{A}}{1-\text{x}}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow\frac{1}{(1-\text{x})(\text{x}^2+1)}=\frac{\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(1-\text{x})}{(1-\text{x})(\text{x}^2+1)}$
$\Rightarrow1=\text{Ax}^2+\text{A}+\text{Bx}-\text{Bx}^2+\text{C}-\text{Cx}$
$\Rightarrow1=(\text{A}-\text{B})\text{x}^2+(\text{B}-\text{C})\text{x}+\text{A}+\text{C}$
Equating coefficient of like terms.
$\text{A}-\text{B}=0\ ...(1)$
$\text{B}-\text{C}=0\ ...(2)$
$\text{A}+\text{C}=1\ ...(3)$
Solving (1), (2) and (3), we get
$\text{A}=\frac{1}{2},\text{B}=\frac{1}{2},\text{C}=\frac{1}{2}$
$\therefore\frac{1}{(1-\text{x})(\text{x}^2+1)}=\frac{1}{2(1-\text{x})}+\frac{\frac{\text{x}}{2}+\frac{1}{2}}{\text{x}^2+1}$
$\int\frac{3\text{dx}}{(1-\text{x})(\text{x}^2+1)}=\frac{3}{2}\int\frac{\text{dx}}{1-\text{x}}+\frac{3}{2}\int\frac{\text{x dx}}{\text{x}^2+1}+\frac{3}{2}\int\frac{\text{dx}}{\text{x}^2+1}$
Putting $\text{x}^2+1=\text{t}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\frac{3}{2}\int\frac{\text{dx}}{1-\text{x}}+\frac{3}{4}\int\frac{\text{dt}}{\text{t}}+\frac{3}{2}\int\frac{\text{dx}}{\text{x}^2+1}$
$=\frac{3}{2}\frac{\log|1-\text{x}|}{-1}+\frac{3}{4}\log|\text{t}|+\frac{3}{2}\times\tan^{-1}\text{x}+\text{C}$
$=\frac{-3}{2}\log|1-\text{x}|+\frac{3}{4}\log|1+\text{x}^2|+\frac{3}{2}\tan^{-1}\text{x}+\text{C}$
$=\frac{-3}{4}\times2\log|1-\text{x}|+\frac{3}{4}\log|1+\text{x}^2|+\frac{3}{2}(2\tan^{-1}\text{x})+\text{C}$
$=\frac{3}{4}\big[\log|1+\text{x}^2|-\log|(1-\text{x}^2)|\big]+\frac{3}{4}(2\tan^{-1}\text{x})+\text{C}$
$=\frac{3}{4}\Big[\log\Big|\frac{1+\text{x}^2}{(1-\text{x})^2}\Big|+2\tan^{-1}(\text{x})\Big]+\text{C}$