$\int\frac{(2x^2+3)dx}{(x^2-1)(x^2+4)}=k\log(\frac{x+1}{x-1})+k(\tan^{-1}(\frac{x}{2}))$, then k equals:
Answer & explanation
Correct answer: option 2
$\int\frac{2x^2+3dx}{(x^2-1)(x^2+4)}=\int\frac{1}{x^2-1}+\frac{1}{x^2+4}dx$
$=\frac{1}{2}\log\left|\frac{x+1}{x-1}\right|+\frac{1}{2}\tan^{-1}\frac{x}{2}+C$
$⇒k=\frac{1}{2}$ On comparison