Integrate the function: $\frac{5 x}{(x+1)(x^{2}+9)}$

Let $\frac{5 x}{(x+1)(x^{2}+9)} = \frac{A}{(x+1)} + \frac{Bx+C}{(x^{2}+9)}$ ........$(1)$
$\Rightarrow 5x = A(x^{2}+9) + (Bx+C)(x+1)$
$\Rightarrow 5x = Ax^{2} + 9A + Bx^{2} + Bx + Cx + C$
Equating the coefficients of $x^{2}, x,$ and the constant term,we obtain:
$A+B = 0$
$B+C = 5$
$9A+C = 0$
On solving these equations,we obtain $A = -\frac{1}{2}, B = \frac{1}{2},$ and $C = \frac{9}{2}$.
From equation $(1)$,we have:
$\frac{5x}{(x+1)(x^{2}+9)} = -\frac{1}{2(x+1)} + \frac{\frac{x}{2} + \frac{9}{2}}{(x^{2}+9)}$
$\int \frac{5x}{(x+1)(x^{2}+9)} dx = \int \left\{ -\frac{1}{2(x+1)} + \frac{x+9}{2(x^{2}+9)} \right\} dx$
$= -\frac{1}{2} \log |x+1| + \frac{1}{2} \int \frac{x}{x^{2}+9} dx + \frac{9}{2} \int \frac{1}{x^{2}+9} dx$
$= -\frac{1}{2} \log |x+1| + \frac{1}{4} \int \frac{2x}{x^{2}+9} dx + \frac{9}{2} \int \frac{1}{x^{2}+9} dx$
$= -\frac{1}{2} \log |x+1| + \frac{1}{4} \log (x^{2}+9) + \frac{9}{2} \cdot \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right) + C$
$= -\frac{1}{2} \log |x+1| + \frac{1}{4} \log (x^{2}+9) + \frac{3}{2} \tan^{-1} \left( \frac{x}{3} \right) + C$