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.

\begin{align}
\int \frac{x + 9}{x^3 + 9x} dx &= \int \frac{x}{(x^2 + 9)x} dx + \int \frac{9}{(x^2 + 9)x} dx 
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&= \int \frac{dx}{x^2 + 9}  + \int \left(\frac{A}{x} + \frac{Bx+C}{x^2 + 9}\right) dx 
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&= \int \frac{dx}{x^2 + 9}  + \int \left(\frac{1}{x} + \frac{-x}{x^2 + 9}\right) dx 
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&= \frac{1}{9}\int \frac{dx}{\left(x/3\right)^2 + 1}  + \int \frac{dx}{x}  - \int\frac{x}{x^2 + 9}dx 
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&= \frac{1}{3}\arctan(x/3)  + \ln x  - \frac{1}{2}\ln(x^2+9) + c
\end{align}
$$\int \frac{x + 9}{x^3 + 9x} dx = k\arctan(mx) + n\ln (x) + p \ln (x^2 + 9) + c$$
$$\frac{m+n}{k+p} = \frac{1/3+1}{1/3-1/2}=\frac{4/3}{-1/6}=-8$$
