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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the functions\[\frac{4x+1}{\sqrt{2x^2+x-3}}\]

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Toolbox:
  • $\int\frac{(px+q)}{\sqrt{ax^2+bx+c}}dx.$,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.
Given:$\int \frac{4x+1}{\sqrt{2x^2+x-3}}dx.$
 
Let 4x+1=A.$\frac{d}{dx}(2x^2+x-3)+B.$
 
$\Rightarrow 4x+1=A(4x+1)+B.$
 
$\Rightarrow 4x+1=4Ax+A+B.$
 
On equating the coefficients we get,
 
4A=4$\Rightarrow A=1.$
 
A+B=1$\Rightarrow B=0.$
 
On substituting for A and B we get
 
Therefore $4x+1=1(4x+1)+0.$
 
Hence $I=\int\frac{4x+1}{\sqrt{2x^2+x-3}}$.
 
Let $2x^2+x-3=t.$
 
On differentiating we get,
 
Therefore $(4x+1)dx=dt.$
 
On substituting we get,
 
$I=\int\frac{dt}{\sqrt t}.$
 
On integrating we get,
 
$\;\;\;=\frac{t^\frac{1}{2}}{\frac{1}{2}}+c.$
 
$\;\;\;=2\sqrt t+c.$
 
Substituting for t we get,
 
$\;\;\;=2\sqrt{2x^2+x-3}+c.$

 

answered Feb 4, 2013 by sreemathi.v
 
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