Question

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

• A. $\frac{1}{5}\log \left(x-1\right|+\frac{1}{10}\log (x^2+4)+\frac{2}{5}{\tan }^{-1}\left(\frac{x}{2}\right)+C$
• B. $\frac{1}{5}\log \left(x-1\right|-\frac{1}{10}\log (x^2+4)+\frac{2}{5}{\tan }^{-1}\left(\frac{x}{2}\right)+C$
• C. $-\frac{1}{5}\log \left(x-1\right|-\frac{1}{10}\log (x^2+4)+\frac{2}{5}{\tan }^{-1}\left(\frac{x}{2}\right)+C$
• D. $\frac{1}{5}\log \left(x-1\right|-\frac{1}{10}\log (x^2+4)-\frac{2}{5}{\tan }^{-1}\left(\frac{x}{2}\right)+C$

Answer 1

The correct option is B $\frac{1}{5}\log \left(x-1\right|-\frac{1}{10}\log (x^2+4)+\frac{2}{5}{\tan }^{-1}\left(\frac{x}{2}\right)+C$

We can write the integrand as:
$\mathrm{let},\frac{x}{(x-1)(x^2+4)}=\frac{A}{x-1}+\frac{\mathrm{Bx}+c}{x^2+4}\mathrm{Solving}\mathrm{this},\mathrm{we}\mathrm{get}A=\frac{1}{5},B=-\frac{1}{5}\mathrm{and}C=\frac{4}{5}.$
Now, substituting the values of $A,B\mathrm{and}C$,$\frac{x}{(x-1)(x^2+4)}=\frac{1}{5(x-1)}+\frac{-\frac{1}{5}x+\frac{4}{5}}{x^2+4}=\frac{1}{5(x-1)}-\frac{1}{5}\frac{(x-4)}{(x^2+4)}$
Or, we can write
$I=\frac{1}{5}\int \frac{dx}{x-1}-\frac{1}{5}\int \frac{x-4}{x^2+4}dx=\frac{1}{5}\int \frac{dx}{x-1}-\frac{1}{10}\int \frac{2x}{x^2+4}dx+\frac{4}{5}\int \frac{1}{x^2+4}dx=\frac{1}{5}\log \left(x-1\right|-\frac{1}{10}\log \left(x^2+4\right|+\frac{4}{5}\times \frac{1}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+C=\frac{1}{5}\log \left(x-1\right|-\frac{1}{10}\log (x^2+4)+\frac{2}{5}{\tan }^{-1}\left(\frac{x}{2}\right)+C(\because (x^2+4|=x^2+4)$