Browse Questions

# Integrate the function $\frac{\large x^2}{(\large 2+3x^3)^3}$

$\begin{array}{1 1} \frac{-1}{18(2+3x^3)^2}+c. \\ \frac{1}{18(2+3x^3)^3}+c. \\ \frac{-1}{18(2+3x^3)^3}+c. \\ \large \frac{1}{9(2+3x^3)^2}+c.\end{array}$

Toolbox:
• Method of substitution :
• Given f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t).
• Consider $I=\int f(x)dx.$
• Put x=g(t) so that $\frac{dx}{dt}=g'(t).$
• dx=g'(t)dt.
• Thus $I=\int f(g(t).g'(t))dt.$
Given $I=\int\frac{x^2}{(2+3x^3)^3}dx$.

Let $(2+3x^3)=t$.

$\;\;\;9x^2dx=dt$.

$\;\;\;x^2dx=\frac{dt}{9}$.

Substituting for x and dx we get,

$I=\int\frac{dt}{9t^3}=\frac{1}{9}\int\frac{dt}{t^3}$.

On integrating on both sides we get,

$\;\;\;=\frac{1}{9}\begin{bmatrix}\frac{t^{-2}}{-2}\end{bmatrix}+c.$

$\;\;\;=\frac{1}{18}[\frac{1}{t^2}+c.$

Substituting back for t we get,

$\;\;\;=\frac{-1}{18(2+3x^3)^2}+c.$

Thus $\int\frac{x^2}{(2+3x^3)^3}dx=\frac{-1}{18(2+3x^3)^2}+c.$