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

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  • $(i)\int\frac{(px+q)}{\sqrt{ax^2+bx+c}}dx.$,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)$\int\frac{dx}{\sqrt{x^2-a^2}}=log\mid x+\sqrt{x^2-a^2}\mid.$
Given:$I=\int \frac{x+2}{\sqrt{x^2-1}}dx.$
Let $x+2=A\frac{d}{dx}(x^2-1)+B.$
Equating the coefficients,
I=2A$\Rightarrow A=\frac{1}{2}.$
Let $x^2-1=t.$
On differentiating we get,
Therefore $2xdx=dt.\Rightarrow xdx=dt.$
On substituting we get,
$I=\frac{1}{2}\int\frac{dt}{\sqrt t}+2\int\frac{dx}{\sqrt{x^2-1}}.$
On integrating we get,
$\;\;\;=\frac{1}{2}\big(\frac{t^\frac{1}{2}}{\frac{1}{2}}\big)+2log\mid x+\sqrt{x^2-1}\mid+c.$
$\;\;\;=\sqrt t+2log\mid x+\sqrt{x^2-1}\mid+c.$
Substituting for t we get,
$\;\;\;=\sqrt{x^2-1}+2log\mid x+\sqrt{x^2-1}\mid+c.$


answered Feb 4, 2013 by sreemathi.v