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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $(3x-2)\sqrt{x^2+x+1}$

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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 (3x-2)\sqrt {x^2+x+1}dx.$
 
Let $x^2+x+1=t $.
 
Hence$ (2x+1)dx=dt.$
 
Now we can Substitute for $ t $and $dt$ in $ I$ we get,
 
$I=\int \large\frac{3}{2}$$(2x-\large\frac{4}{3}$$)\sqrt{x^2+x+1}dx.$
 
$I= \large\frac{3}{2}$$\int(2x+1-\large\frac{7}{3}$$)\sqrt{x^2+x+1}dx.$
$I= \large\frac{3}{2}$$\big[\int(2x+1)\sqrt{x^2+x+1}dx.-\int\large\frac{7}{3}$$\sqrt{x^2+x+1}dx\big]$
$\;\;\;=\large\frac{3}{2}$$\big[\int (t)^\frac{1}{2},dt-\large\frac{7}{3}\int\sqrt{x^2+x+1}dx\big]$
 
On integrating we get,
 
$\;\;\;=t^{3/2}-\large\frac{7}{2}\bigg[\large\frac{x+1/2}{2}$$\sqrt{x^2+x+1}+\large\frac{3}{8}$$log|x+1/2+\sqrt{x^2+x+1}|\bigg]$
 
 
Substituting back for t we get,
 
 
$I=(x^2+x+1)^{3/2}-\large\frac{7}{2}\bigg[\large\frac{2x+1}{4}$$\sqrt{x^2+x+1}+\large\frac{3}{8}$$log|x+1/2+\sqrt{x^2+x+1}|\bigg]$
answered Mar 21, 2014 by rvidyagovindarajan_1
edited Mar 21, 2014 by rvidyagovindarajan_1
 

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