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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the functions\[\frac{1}{9x^2+6x+5}\]

$\begin{array}{1 1} \frac{1}{6} \tan^{-1}\big(\frac{3x+1}{2}\big)+c \\ \frac{1}{3} \tan^{-1}\big(\frac{3x+1}{2}\big)+c \\ \frac{1}{6} \tan^{-1}\big(\frac{x+1}{2}\big)+c \\ \frac{1}{6} \tan^{-1}\big(2\frac{3x+1}{3}\big)+c \end{array} $

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1 Answer

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  • $\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\big(\frac{x}{a}\big)+c.$
Given:$I=\int\frac{dx}{9x^2+6x+5}.$
 
$9x^2+6x+5$ can be written as $(3x+1)^2+2^2.$
 
Hence $I=\int\frac{dx}{(3x+1)^2+2^2}.$
 
Let 3x+1=t.
 
On differentiating we get,
 
Therefore 3dx=dt $\Rightarrow dx=\frac{dt}{3}.$
 
On substituting this we get,
 
$I=\int\frac{dt/3}{(t^2)+2^2}=\frac{1}{3}\int\frac{dt}{t^2+2^2}.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{3}.\frac{1}{2}\tan^{-1}\big(\frac{t}{2}\big)+c.$
 
Substituting for t we get,
 
$\;\;\;=\frac{1}{6}\tan^{-1}\big(\frac{3x+1}{2}\big)+c.$

 

answered Feb 4, 2013 by sreemathi.v
 
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