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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $(4x+2)\sqrt{x^2+x+1}$

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Toolbox:
  • Method of substitution:
  • Given $\int f(x)dx$ 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}{dx}=g'(t).
  • dx=g'(t)dt.
  • Thus $ I=\int f(g(t).g'(t))dt.$
Given $ I=\int (4x+2)\sqrt {x^2+x+1}dx.$
 
Let $x^2+x+1=t $.
 
Hence 2x+1=dt.
 
Now we can Substitute for t and dt in I we get,
 
$I=\int 2(2x+1)\sqrt{x^2+x+1}dx.$
 
$\;\;\;=\int 2(t)^\frac{1}{2},dt$
 
On integrating we get,
 
$\;\;\;=2\frac{t^\frac{3}{2}}{\frac{3}{2}}+c$.
 
$\;\;\;=2\times\frac{2}{3}(t^\frac{3}{2})+c.$
 
Substituting back for t we get,
 
$\;\;\;=\frac{4}{3}[(x^2+x+1)^\frac{3}{2}]+c.$
 
Hence $ \int (4x+2)\sqrt {x^2+x+1}dx=\frac{4}{3}[(x^2+x+1)^\frac{3}{2}]+c.$

 

answered Jan 28, 2013 by sreemathi.v
 
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