# Show that

Question:

$\frac{5 x}{(x+1)\left(x^{2}+9\right)}$

Solution:

Let $\frac{5 x}{(x+1)\left(x^{2}+9\right)}=\frac{A}{(x+1)}+\frac{B x+C}{\left(x^{2}+9\right)}$    ...(1)

$\Rightarrow 5 x=A\left(x^{2}+9\right)+(B x+C)(x+1)$

$\Rightarrow 5 x=A x^{2}+9 A+B x^{2}+B x+C x+C$

Equating the coefficients of $x^{2}, x$, and constant term, we obtain

$A+B=0$

$B+C=5$

$9 A+C=0$

On solving these equations, we obtain

$A=-\frac{1}{2}, B=\frac{1}{2}$, and $C=\frac{9}{2}$

From equation (1), we obtain

$\frac{5 x}{(x+1)\left(x^{2}+9\right)}=\frac{-1}{2(x+1)}+\frac{\frac{x}{2}+\frac{9}{2}}{\left(x^{2}+9\right)}$

$\int \frac{5 x}{(x+1)\left(x^{2}+9\right)} d x=\int\left\{\frac{-1}{2(x+1)}+\frac{(x+9)}{2\left(x^{2}+9\right)}\right\} d x$

$=-\frac{1}{2} \log |x+1|+\frac{1}{2} \int \frac{x}{x^{2}+9} d x+\frac{9}{2} \int \frac{1}{x^{2}+9} d x$

$=-\frac{1}{2} \log |x+1|+\frac{1}{4} \int \frac{2 x}{x^{2}+9} d x+\frac{9}{2} \int \frac{1}{x^{2}+9} d x$

$=-\frac{1}{2} \log |x+1|+\frac{1}{4} \log \left|x^{2}+9\right|+\frac{9}{2} \cdot \frac{1}{3} \tan ^{-1} \frac{x}{3}$

$=-\frac{1}{2} \log |x+1|+\frac{1}{4} \log \left(x^{2}+9\right)+\frac{3}{2} \tan ^{-1} \frac{x}{3}+C$