Question

Solve:$\int \frac{x^2}{1+2x}dx=?$

Answer 1

$\int \frac{x^2}{1+2x}dx=\frac{1}{2}\int \frac{2x^2+x}{2x+1}dx-\frac{1}{2}\int \frac{x}{1+2x}dx$

(Adding & subtracting $\frac{x}{(1+2x)}$after multiplying & dividing by 2)

$=\frac{1}{2}\int \frac{x(2x+1)}{(2x+1)}dx-[\frac{1}{2}.\frac{1}{2}\int \frac{2x+1}{2x+1}dx-\frac{1}{2}.\frac{1}{2}\int \frac{1}{1+2x}dx]$

(Adding & subtracting $\frac{1}{(1+2x)}$ after multiplying & dividing $2^{nd}$ term by $2$)

$=\frac{1}{2}\int xdx-[\frac{1}{4}\int 1dx-\frac{1}{4}\int \frac{dx}{1+2x}]$

$=\frac{x^2}{2.2}-\frac{x}{4}+\frac{1}{4}.\frac{1}{2}\int \frac{2dx}{1+2x}+C^{\prime }$

$=\frac{x^2}{4}-\frac{x}{4}+\frac{1}{8}\log |1+2x|+C$ $($ Here $f\left(x\right)=1+2x$ & we know $\int \frac{f^1\left(x\right)}{f\left(x\right)}dx=\log |f\left(x\right)|+C)$