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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the functions\[\frac{6x+7}{\sqrt{(x-5)(x-4)}}\]

$\begin{array}{1 1} 6(\sqrt{x^2-9x+2})+34[\log|(x-9/2)+\sqrt{(x-5)(x-4)}|]+c \\ 6(\sqrt{x^2-9x+2})-34[\log|(x-9/2)+\sqrt{(x-5)(x-4)}|]+c \\ 6(\sqrt{x^2-9x+2})-17[\log|(x-9/2)-\sqrt{(x-5)(x-4)}|]+c \\ 6(\sqrt{x^2-9x+2} -17 [\log|(x+9/2)-\sqrt{(x-5)(x-4)}|]+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:$I=\int \frac{6x+7}{(x-5)(x-4)}dx=\int\frac{6x+7}{\sqrt{x^2-9x+20}}.$
 
We can write,
 
6x+7=$A\frac{d}{dx}(x^2-9x+20)+B.$
 
$\;\;\;=A(2x-9)+B.$
 
Now equating the coefficients we get,
 
6=2A$\Rightarrow A=3.$
 
$7=-9A+B\Rightarrow B=34.$
 
Hence A=3 and B=34.
 
$I=\int\frac{3(2x-9)+34}{\sqrt{x^2-9x+20}}$.
 
On separating we can write,
 
$I=3\int\frac{(2x-9)dx}{\sqrt{x^2-9x+20}}+34\int\frac{dx}{\sqrt{x^2-9x+20}}.$
 
Let $x^2-9x+20=t.$ and $x^2-9x+20=(x-\frac{9}{2})^2-\frac{1}{4}=(x-\frac{9}{2})^2-(\frac{1}{2})^2$.
 
On differentiating
 
$(2x-9)dx=dt.$
 
Therefore $I=3\int\frac{dt}{\sqrt t}+34\int\frac{dx}{\sqrt{(x-\frac{9}{2}}})^2-(\frac{1}{2})^2.$
 
On integrating we get,
 
$\;\;\;=3\begin{bmatrix}\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\end{bmatrix}+34[log|(x-9/2)+\sqrt{(x-5)(x-4)}|]$.
 
$\;\;\;=3\times 2(\sqrt t)+34log|(x-9/2)+\sqrt{(x-5)(x-4)}+c.$
 
Substituting for t we get,
 
$\int\frac{6x+7}{\sqrt{(x-5)(x-4)}}dx=6(\sqrt{x^2-9x+2})+34[log|(x-9/2)+\sqrt{(x-5)(x-4)}|]+c.$

 

answered Feb 4, 2013 by sreemathi.v
edited Feb 4, 2013 by sreemathi.v
 
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