Question

If $\int \frac{dx}{(x^2+1{)}^2}=\frac{A}{148}ta{n}^{-1}x+\frac{1}{2}\frac{x}{x^2+1}+C$ then A is equal to

Answer 1

Let $I=\int \frac{1}{{(x^2+1)}^2}dx$
Substituting $x=\tan t\Rightarrow dx={\sec }^{2}tdt$
$I=\int {\cos }^{2}tdt=\frac{1}{2}\int (\cos 2t+1)dt=\frac{t}{2}+\frac{\sin 2t}{4}=\frac{x^2{\tan }^{-1}x+x+{\tan }^{-1}x}{2x^2+2}+c=\frac{1}{2}(\frac{x}{x^2+1}+{\tan }^{-1}x)+c$
Therefore, $A=74$