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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\frac{\large x^2}{(\large 2+3x^3)^3}$

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

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1 Answer

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

 

answered Jan 28, 2013 by sreemathi.v
 
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