Question

If $\int \frac{dx}{(x^2+1)(x^2+4)}=Kta{n}^{-1}x+Lta{n}^{-1}\frac{x}{2}+c$ where c is arbitrary constant, then K + L =

• A. $\frac{1}{6}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $0$

Answer 1

The correct option is A $\frac{1}{6}$

$\frac{1}{(x^2+1)(x^2+4)}=\frac{1}{3}.\frac{1}{(x^2+1)}-\frac{1}{3}.\frac{1}{(x^2+4)}$
$\therefore \int \frac{dx}{(x^2+1)(x^2+4)}=\frac{1}{3}ta{n}^{-1}x-\frac{1}{6}ta{n}^{-1}\left(\frac{x}{2}\right)+C$
Comparing with R.H.S., we have $k=\frac{1}{3},L=\frac{-1}{6}$
$\therefore K+L=\frac{1}{6}$