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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the functions\[\frac{5x+3}{\sqrt{x^2+4x+10}}\]

$\begin{array}{1 1} 5\log|x^2+4x+10|-7\log|(x+2)+\sqrt{x^2+4x+10}|+c \\5\log|x^2+4x+10|+7\log|(x+2)+\sqrt{x^2+4x+10}|+c \\ 5\log|x^2+4x+10|+7\log|(x+2)-\sqrt{x^2+4x+10}|+c \\ 5\log|x^2+4x+10|-7\log|(x+2)-\sqrt{x^2+4x+10}|+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)$\int\frac{dx}{\sqrt{x^2+a^2}}=log\mid x+\sqrt{x^2+a^2}\mid+c.$
Given:$I=\int \frac{5x+3}{x^2+4x+10}dx.$
 
Let $5x+3=A\frac{d}{dx}(x^2+4x+10)+B.$
 
$\quad 5x+3=A(2x+4)+B.$
 
Now equating the coefficients of x we get,
 
5=2A$\Rightarrow A=\frac{5}{2}.$
 
equating the coefficients we get,
 
3=4A+B$\Rightarrow B=-7\;and\;A=\frac{5}{2}.$
 
Hence A=5/2 and B=-7.
 
$I=\int\frac{\frac{5}{2}(2x+4)-7}{\sqrt{x^2+4x+10}}dx.$
 
On separating the terms we can write,
 
$I=\frac{5}{2}\int\frac{2x+42}{\sqrt{x^2+4x+10}}-7\int\frac{dx}{\sqrt{x^2+4x+10}}.$
 
Let $x^2+4x+10=t.$
 
Therefore (2x+4)dx=dt.
 
And $x^2+4x+10$ can be written as $(x+2)^2-(\sqrt 6)^2.$
 
On substituting we get,
 
$I=\frac{5}{2}\int\frac{dt}{t}-7\int\frac{dx}{(x+2)^2+(\sqrt 6)^2}.$
 
On integrating we get,
 
$\;\;\;=\frac{5}{2}\begin{bmatrix}\frac{t^{-1/2+1}}{-1/2+1}\end{bmatrix}-7log \mid(x+2)+\sqrt{x^2+4x+10}\mid+c.$
 
Substituting for t we get,
 
$\;\;\;=5log|x^2+4x+10|-7log|(x+2)+\sqrt{x^2+4x+10}|+c.$

 

answered Feb 5, 2013 by sreemathi.v
 
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