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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\frac{\large x}{\large \sqrt{x+4}},x\;>\;0$

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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 $I=\int\frac{x}{\sqrt{x+4}}dx$.
 
Let x+4=t [dx=dt]
 
$\Rightarrow x=t-4$.
 
Now substituting for x and dx we get,
 
$\int\frac{x}{\sqrt {x+4}}dx=\int \frac{(t-4)}{\sqrt t}dt$.
 
$\;\;\;\qquad\quad\;=\int \sqrt tdt-\int\frac{4}{\sqrt t}dt.$
 
On integrating we get,
 
$\;\;\;\qquad\quad\;=\begin{bmatrix}\frac{t^\frac{3}{2}}{\frac{3}{2}}\end{bmatrix}-4\begin{bmatrix}\frac{t^\frac{1}{2}}{\frac{1}{2}}\end{bmatrix}+c.$
 
$\;\;\;\qquad\quad\;=\frac{2}{3}[t^\frac{3}{2}]-8(t^\frac{1}{2})+c.$
 
Substituting back for t we get,
$\;\;\;\qquad\quad\;=\frac{2}{3}\begin{bmatrix}(x+4)^\frac{3}{2}\end{bmatrix}-4\begin{bmatrix}(x+4)^\frac{1}{2}\end{bmatrix}+c.$
 
On further simplifying,
$\;\;\;\qquad\quad\;=\frac{2}{3}[(x+4)^\frac{1}{2}](x+4-12)+c.$
 
$\;\;\;\qquad\quad\;=\frac{2}{3}\sqrt{(x+4)}(x-8)+c.$
 
Hence $\int\frac{x}{\sqrt{x+4}}dx=\frac{2}{3}\sqrt{(x+4)}(x-8)+c.$

 

answered Jan 28, 2013 by sreemathi.v
 
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