Integrals — Class 12 Maths Solution

: Let$I = \int {\cfrac{{{x^2}}}{{{{\left( {2 + 3{x^3}} \right)}^3}}}dx}$

Put $2 + 3{x^2} = t$ $\Rightarrow$ $9{x^2}dx = dt$

$\therefore$ $I = \cfrac{1}{9}\int {\cfrac{{9{x^2}\,dx}}{{{{\left( {2 + 3{x^3}} \right)}^3}}} = \cfrac{1}{9}\int {\cfrac{{dt}}{{{t^3}}} = \cfrac{1}{9}\int {{t^{ - 3}}} dt} }$

$= \cfrac{1}{9}\cfrac{{{t^{ - 2}}}}{{\left( { - 2} \right)}} + C = \cfrac{1}{{18}}{t^{ - 2}} + C = - \cfrac{1}{{18{{\left( {2 + 3{x^3}} \right)}^2}}} + C$