Question

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

Answer 1

Let $\frac{5x}{(x+1)(x^2+9)}=\frac{A}{(x+1)}+\frac{Bx+C}{(x^2+9)}$ ......... (1)
$\Rightarrow 5x=A(x^2+9)+(Bx+C)(x+1)$
$\Rightarrow 5x=A{x}^2+9A+B{x}^2+Bx+Cx+C$
Equating the coefficients of $x^2,x$, and 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 obtain
$\frac{5x}{(x+1)(x^2+9)}=\frac{-1}{2(x+1)}+\frac{\frac{x}{2}+\frac{9}{2}}{(x^2+9)}$
$\int \frac{5x}{(x+1)(x^2+9)}dx=\int \{\frac{-1}{2(x+1)}+\frac{(x+9)}{2(x^2+9)}\}dx$
$=-\frac{1}{2}\log |x+1|+\frac{1}{2}\int \frac{x}{x^2+9}dx+\frac{9}{2}\int \frac{1}{x^2+9}dx$
$=-\frac{1}{2}\log |x+1|+\frac{1}{4}\int \frac{2x}{x^2+9}dx+\frac{9}{2}\int \frac{1}{x^2+9}dx$
$=-\frac{1}{2}\log |x+1|+\frac{1}{4}\log |x^2+9|+\frac{9}{2}\cdot \frac{1}{3}{\tan }^{-1}\frac{x}{3}$
$=-\frac{1}{2}\log |x+1|+\frac{1}{4}\log (x^2+9)+\frac{3}{2}{\tan }^{-1}\frac{x}{3}+C$