Question

If $\int \frac{dx}{(x^2+4)(x^2+7)}=K{\tan }^{-1}\frac{x}{\sqrt{7}}+L{\tan }^{-1}\frac{x}{2}+C$, then

• A. $K=-\frac{1}{\sqrt{7}}$
• B. $L=\frac{1}{6}$
• C. $L=-\frac{1}{3}$
• D. $K=-\frac{1}{3\sqrt{7}}$

Answer 1

Answer: (B), (D)

The correct options are
B $L=\frac{1}{6}$
D $K=-\frac{1}{3\sqrt{7}}$

$\int \frac{1}{(x^2+4)(x^2+7)}dx=\int \frac{1}{3}(\frac{1}{x^2+4}-\frac{1}{x^2+7})dx$

$=-\frac{1}{3\sqrt{7}}{\tan }^{-1}\frac{x}{\sqrt{7}}+\frac{1}{6}{\tan }^{-1}\frac{x}{2}+c$