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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the functions\[\frac{x+3}{x^2-2x-5}\]

$\begin{array}{1 1} \frac{1}{2}\log|x^2-2x-5|+\frac{2}{\sqrt 6}\log\mid\frac{x-1-\sqrt 6}{x-1+\sqrt 6}\mid+c \\ \frac{1}{2}\log|x^2-2x-5|-\large \frac{2}{\sqrt 6}\log\mid\frac{x-1-\sqrt 6}{x-1+\sqrt 6}\mid+c \\ \frac{1}{2}\log|x^2-2x-5|+\large \frac{1}{\sqrt 3}\log\mid\frac{x-1-\sqrt 6}{x-1+\sqrt 6}\mid+c \\ \frac{1}{2}\log|x^2-2x-5|- \frac{2}{\sqrt 3}\log\mid\frac{x-1-\sqrt 6}{x-1+\sqrt 6}\mid+c \end{array} $

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1 Answer

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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)$\int\frac{dx}{\sqrt{x^2-a^2}}=\frac{1}{2a}log\mid\frac{x-a}{x+a}\mid+c.$
Given:$I=\int \frac{x+3}{x^2-2x-5}dx.$
 
Let $x+3=A\frac{d}{dx}(x^2-2x-5)+B.$
 
$\;\;\;\;\quad =A(2x-2)+B.$
 
Now equating the coefficients of x we get,
 
1=2A$\Rightarrow A=\frac{1}{2}.$
 
equating the coefficients we get,
 
3=-2A+B$\Rightarrow B=4\;and\;A=\frac{1}{2}.$
 
Hence A=1/2 and B=4.
 
$I=\int\frac{\frac{1}{2}(2x-2)+4}{x^2-2x-5}dx.$
 
On separating the terms we can write,
 
$I=\frac{1}{2}\int\frac{2x-2}{x^2-2x-5}+4\int\frac{dx}{x^2-2x-5}.$
 
Let $x^2-2x-5=t.$
 
Therefore (2x-2)dx=dt.
 
And $x^2-2x-5$ can be written as $(x-1^2-(\sqrt 6)^2.$
 
On substituting we get,
 
$I=\frac{1}{2}\int\frac{dt}{t}+4\int\frac{dx}{(x-1)^2-(\sqrt 6)^2}.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{2}log t+4.\frac{1}{2\sqrt 6}log\mid\frac{x-1-\sqrt 6}{x-1+\sqrt 6}\mid+c.$+c.$
 
Substituting for t ,
 
Hence $\int\frac{x+3}{x^2-2x-5}dx=\frac{1}{2}log|x^2-2x-5|+\frac{2}{\sqrt 6}log\mid\frac{x-1-\sqrt 6}{x-1+\sqrt 6}\mid+c.$

 

answered Feb 5, 2013 by sreemathi.v
 
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