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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the functions\[\frac{x+2}{\sqrt{x^2+2x+3}}\]

$\begin{array}{1 1} \sqrt{x^2+2x+3}+\log|(x+1)+\sqrt{x^2+2x+3}|+c \\ \sqrt{x^2+2x+3}-\log|(x+1)-\sqrt{x^2+2x+3}|+c \\ \sqrt{x^2+2x+3}-\log|(x+1)+\sqrt{x^2-2x-3}|+c \\ \sqrt{x^2-2x-3}-\log|(x+1)+\sqrt{x^2+2x+3}|+c \end{array} $

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Toolbox:
  • $(i)\int\frac{(px+q)}{\sqrt{ax^2+bx+c}}dx.$,where p,q,a,b,c are constants,we are to find real numbers such that \[px+q=A\frac{d}{dx}(ax^2+bx+c)+B.\]
  • Therefore $px+q=A(2ax+b)+B.$
  • To determine A and B we equate the coefficients from both sided,the coefficients of x and the constant terms.A and B are thus obtained and hence the integral is reduced to one of the known forms:
  • (ii)$\int\frac{dx}{\sqrt{x^2+a^2}}=log|x+\sqrt{x^2+a^2}|+c.$
Given:$I=\int \frac{x+2}{\sqrt{x^2+2x+3}}dx.$
 
Let x+2=$A\frac{d}{dx}(x^2+2x+3)+B.$
 
$\;\;x+2=A(2x+2)+B.$
 
Now equating the coefficients of x we get,
 
1=2A$\Rightarrow A=\frac{1}{2}.$
 
equating the coefficients we get,
 
1=2A+B$\Rightarrow B=1$.
 
Hence A=1/2 and B=1.
 
$I=\int\frac{\frac{1}{2}(2x+2)+1}{\sqrt{x^2+2x+3}}dx.$
 
On separating the terms we can write,
 
$I=\frac{1}{2}\int\frac{2x+2}{\sqrt{x^2+2x+3}}+\int\frac{dx}{\sqrt{x^2+2x+3}}.$
 
If $x^2+2x+3=t.$
 
Therefore $(2x+2)dx=dt.$ and $\sqrt{x^2+2x+3}=\sqrt{(x+1)^2+(\sqrt 2)^2}.$
 
Therefore $I=\frac{1}{2}\int\frac{dt}{\sqrt t}+\int\frac{dx}{\sqrt{(x+1)^2+(\sqrt 2)^2}}.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{2}\begin{bmatrix}\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\end{bmatrix}+log|(x-1)+\sqrt{x^2+2x+3}|+c.$
 
$\;\;\;=\frac{1}{2}\big(\frac{\sqrt t}{1/2}\big)+log|(x+1)+\sqrt{x^2+2x+3}|+c.$
 
Substituting for t ,
 
Hence $\int\frac{x+2}{\sqrt{x^2+2x+3}}dx=\sqrt{x^2+2x+3}+log|(x+1)+\sqrt{x^2+2x+3}|+c.$

 

answered Feb 5, 2013 by sreemathi.v
 
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