$\int\frac{(2x^2+3)dx}{(x^2-1)(x^2+4)}=k\log(\frac{x+1}{x-1})+k(\tan^{-1}(\frac{x}{2}))$, then k equals:

log 2 1/2 1/3 $\frac{1}{\log 2}$

1/2

$\int\frac{2x^2+3dx}{(x^2-1)(x^2+4)}=\int\frac{1}{x^2-1}+\frac{1}{x^2+4}dx$ $=\frac{1}{2}\log\left|\frac{x+1}{x-1}\right|+\frac{1}{2}\tan^{-1}\frac{x}{2}+C$ $⇒k=\frac{1}{2}$ On comparison