Multiple choice question of indefinite integral, $\int \frac{x + 9}{x^3 + 9x} dx$. If $\int \frac{x + 9}{x^3 + 9x} dx = k\arctan(mx) + n\ln (x) + p \ln (x^2 + 9) + c$, then $(m+n)/(k+p) = $
(A) 6
(B) -8
(C) -3
(D) 4  
I tried solving it by differentiating the R.H.S. but couldn't arrive at the answer.

Try to separate the denominator like this
$$\frac{x+9}{x^3+9x}=\frac{x+9}{x(x^2+9)}$$
$$\qquad\qquad\ \,=\frac Ax+\frac{Bx+C}{x^2+9}$$
Then construct an equality for $A,B,C$ and we have
$$Ax^2+9A+Bx^2+Cx=x+9$$
Here we get $A=1,\ B=-1,\ C=1$, and that gives us
$$\int\frac{x+9}{x^3+9x}\,dx\ =\ \int\frac1x+\frac1{x^2+9}-\frac x{x^2+9}\,dx$$
Could you continue with this?
$----------$
Note that
\begin{align*}
&\int\frac{x}{x^2+9}\,dx\\
=\ \frac12&\int\frac{du}{u+9}\qquad(u=x^2,\ du=2x\,dx)\\
=\ \frac12&\ln(u+9)+c\\
=\ \frac12&\ln(x^2+9)+c
\end{align*}
