Integrals — Class 12 Maths Solution

: Let$I = \int {x\sqrt {1 + 2{x^2}} } dx$
Put $1 + 2{x^2} = t$ $\Rightarrow$ $4xdx = dt$

$\therefore$ $I = \cfrac{1}{4}\int {\sqrt {1 + 2{x^2}} 4xdx = \cfrac{1}{4}\int {\sqrt t } dt}$

$= \cfrac{1}{4}\cfrac{{{t^{3/2}}}}{{3/2}} + C = \cfrac{1}{6}{t^{3/2}} + C = \cfrac{1}{6}{\left( {1 + 2{x^2}} \right)^{3/2}} + C$