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Q11  Integrate the functions\frac{1}{9 x ^2 + 6x + 5 }

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\frac{1}{9 x ^2 + 6x + 5 }
this denominator can be written as 
9x^2+6x+5=9[x^2+\frac{2}{3}x+\frac{5}{9}]\\=9[(x+\frac{1}{3})^2+(\frac{2}{3})^2] 

Now, 
 \frac{1}{9}\int \frac{1}{(x+\frac{1}{3})^2+(\frac{2}{3})^2}dx =\frac{1}{9} [\frac{3}{2}\tan^{-1}(\frac{(x+1/3)}{2/3})] +C\\=\frac{1}{6} \tan^{-1}(\frac{3x+1}{2})] +C
                                              ......................................by using the form (\int \frac{1}{x^2+a^2}=\frac{1}{a}\tan^{-1}(\frac{x}{a}))
                                                       

Posted by

manish

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