Question

$\int \frac{4x}{(x^2+1)(x^2+3)}dx$ is equal to:

• A. $\log \left[\frac{x^2+1}{x^2+3}\right]+c$
• B. $\log \left[\frac{x^2+3}{x^2+1}\right]+c$
• C. ${\tan }^{-1}x+\left(\frac{1}{\sqrt{3}}\right){\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+c$
• D. $\left(\frac{4}{\sqrt{3}}\right){\tan }^{-1}x{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+c$
• E. $2\log (x^2+1)(x^2+3)+c$

Answer 1

Answer: (A)

Let $I=\int \frac{4x}{(x^2+1)(x^2+3)}dx=\int \frac{2x}{x^2+1}dx$ $-\int \frac{2x}{x^2+3}dx$ Let $x^2+1=u$ and $x^2+3=v$ $\Rightarrow$ $2xdx=du$ and $2xdx=dv$ $\therefore$ $I=\int \frac{du}{u}-\int \frac{du}{v}$ $=logu-logv+c$ $=\log \frac{(x^2+1)}{(x^2+3)}+c$