Question

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

• A. $\frac{1}{2}log(x-1)-\frac{1}{4}log(x^2+1)-\frac{1}{2}{\tan }^{-1}x+c$
• B. $\frac{1}{2}log(x-1)+\frac{1}{4}log(x^2+1)-\frac{1}{2}{\tan }^{-1}x+c$
• C. $\frac{1}{2}log(x-1)-\frac{1}{2}log(x^2+1)-\frac{1}{2}{\tan }^{-1}x+c$
• D. $\frac{1}{2}log(x-1)-\frac{1}{4}log(x^2+1)+{\tan }^{-1}x+c$

Answer 1

The correct option is A $\frac{1}{2}log(x-1)-\frac{1}{4}log(x^2+1)-\frac{1}{2}{\tan }^{-1}x+c$

$\int \frac{1}{(x-1)(x^2+1)}dx=\int (\frac{-1-x}{2(x^2+1)}+\frac{1}{2(x-1)})dx$
$=\frac{1}{2}\int \frac{-1-x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x-1}dx$
$=\frac{1}{2}\int (-\frac{1}{x^2+1}-\frac{x}{x^2+1})dx+\frac{1}{2}\int \frac{1}{x-1}dx$
$=-\frac{1}{2}\int \frac{x}{x^2+1}dx-\frac{1}{2}\int \frac{1}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x-1}dx$
$=-\frac{1}{4}\log (x^2+1)-\frac{1}{2}{\tan }^{-1}x+\frac{1}{2}\log (x+1)$
Hence, option 'A' is correct.