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Integrate the functions\[\frac{4x+1}{\sqrt{2x^2+x-3}}\]

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  • $\int\frac{(px+q)}{\sqrt{ax^2+bx+c}}dx.$,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,
$\;\;\;=2\sqrt t+c.$
Substituting for t we get,


answered Feb 4, 2013 by sreemathi.v