Question

Integrate the following functions w.r.t. x.

$\int \frac{5x}{(x+1)(x^2+9)}dx$

Answer 1

$Let\int \frac{5x}{(x+1)(x^2+9)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+9}\Rightarrow 5x=A(x^2+9)+(Bx+C)(x+1)\Rightarrow 5x=A{x}^2+9A+B{x}^2+Bx+Cx+C\text{On equating the coefficients of}{x}^2\text{,x and constant term on both sides, we get}A+B=0,B+C=5,9A+C=0\text{On solving these equations, we get}A=-\frac{1}{2},B=\frac{1}{2}andC=\frac{9}{2}\therefore \int \frac{5x}{(x+1)(x^2+9)}dx=\int \frac{(-\frac{1}{2})}{(x+1)}dx+\int \frac{\frac{1}{2}x+\frac{9}{2}}{x^2+9}dx=-\frac{1}{2}\int \frac{1}{x+1}dx+\frac{1}{2}\int \left(\frac{x+9}{x^2+9}\right)dx=-\frac{1}{2}log|x+1|+\frac{1}{2}.\frac{1}{2}\int \frac{2x}{x^2+9}dx+\frac{9}{2}\int \frac{1}{x^2+9}dx=-\frac{1}{2}log|x+1|+\frac{1}{4}|x^2+9|+\frac{9}{2}.\frac{1}{3}tan{t}^{-1}\left(\frac{x}{3}\right)+C[\because \int \frac{dx}{a^2+x^2}=\frac{1}{a}ta{n}^{-1}\left(\frac{x}{a}\right)]=-\frac{1}{2}log|x+1|+\frac{1}{4}|x^2+9|+\frac{3}{2}ta{n}^{-1}\left(\frac{x}{3}\right)+C$