Question

Solve

$\int \frac{5x^2+20x+6}{x^3+2x^2+x}dx$

Answer 1

$\int \frac{5x^2+20x+6}{x^3+2x^2+x}=\int \frac{5x^2+20x+6}{x(x^2+2x+1)}dx=\int \frac{5x^2+20x+6}{x{(x+1)}^2}dx$

Using partial cfraction,

Let, $\frac{5x^2+20x+6}{x{(x+1)}^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1{)}^2}=\frac{A(x+1{)}^2+Bx(x+1)+Cx}{x(x+1{)}^2}$

Thus, $5x^2+20x+6=A{x}^2+A+2Ax+B{x}^2+Bx+Cx=(A+B)x^2+(2A+B+C)x+A$

Comparing L.H.S and R.H.S

$A+B=5,2A+B+C=20,A=6$

Substituting $A=6,A+B=5\Rightarrow 6+B=5\Rightarrow B=5-6=-12A+B+C=20\Rightarrow 2\left(6\right)-1+C=20\Rightarrow C=20-11=9$

Substituting values of $A,B$ and $C$

$\frac{5x^2+20x+6}{x{(x+1)}^2}=\frac{6}{x}-\frac{1}{x+1}+\frac{9}{(x+1{)}^2}\int \frac{5x^2+20x+6}{x{(x+1)}^2}dx=6\int \frac{dx}{x}-\int \frac{dx}{x+1}+9\int \frac{dx}{{(x+1)}^2}=6\log \left|x\right|-\log |x+1|-\frac{9}{(x+1)}+C=\log \frac{x^6}{|x+1|}-\frac{9}{x+1}+C$