Question

The value of the integral $\int \frac{x}{(x-1)(x^2+4)}dx$ (where $C$ is integration constant)

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

Answer 1

The correct option is A $\frac{1}{5}\log |x-1|-\frac{1}{10}\log |x^2+4|+\frac{2}{5}{\tan }^{-1}\frac{x}{2}+C$

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

Let $\frac{x}{(x-1)(x^2+4)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+4}\cdots \left(1\right)$
$\Rightarrow x=A(x^2+4)+(Bx+C)(x-1)\cdots \left(2\right)$
Putting $x=1$ in equation $\left(2\right)$, we get $1=5A$
Putting $x=0$ in equation $\left(2\right)$, we get $0=4A-C$
Putting $x=-1$ in equation $\left(2\right)$, we get $-1=5A+2B-2C$

Solving these equations, we obtain $A=\frac{1}{5},B=-\frac{1}{5}$ and $C=\frac{4}{5}$
Substituting the values of $A,B$ and $C$ in equation $\left(1\right)$, we obtain
$\frac{x}{(x-1)(x^2+4)}=\frac{1}{5(x-1)}+\frac{\frac{-1}{5}x+\frac{4}{5}}{x^2+4}$
$=\frac{1}{5(x-1)}-\frac{1}{5}\frac{(x-4)}{(x^2+4)}$
$\Rightarrow I=\frac{1}{5}\int \frac{1}{x-1}dx-\frac{1}{5}\int \frac{x-4}{x^2+4}dx$
$\Rightarrow I=\frac{1}{5}\int \frac{1}{x-1}dx-\frac{1}{10}\int \frac{2x}{x^2+4}dx+\frac{4}{5}\int \frac{1}{x^2+4}dx$
$\Rightarrow I=\frac{1}{5}\log |x-1|-\frac{1}{10}\log |x^2+4|+\frac{4}{5}\times \frac{1}{2}{\tan }^{-1}\frac{x}{2}+C$
$\therefore I=\frac{1}{5}\log |x-1|-\frac{1}{10}\log |x^2+4|+\frac{2}{5}{\tan }^{-1}\frac{x}{2}+C$