Integrals — Class 12 Maths Solution

Let$I = \int {\cfrac{1}{{9{x^2} + 6x + 5}}} = \cfrac{1}{9}\int {\cfrac{{dx}}{{{x^2} + \cfrac{2}{3}x + \cfrac{5}{9}}}}$

$= \cfrac{1}{9}\int {\cfrac{{dx}}{{\left( {{x^2} + \cfrac{2}{3}x + \cfrac{1}{9}} \right) + \left( {\cfrac{5}{9} - \cfrac{1}{9}} \right)}} = \cfrac{1}{9}\int {\cfrac{{dx}}{{{{\left( {x + \cfrac{1}{3}} \right)}^2} + {{\left( {\cfrac{2}{3}} \right)}^2}}}} }$

$= \cfrac{1}{9} \times \cfrac{1}{{2/3}}{\tan ^{ - 1}}\left( {\cfrac{{x + \cfrac{1}{3}}}{{\cfrac{2}{3}}}} \right) + C = \cfrac{1}{6}{\tan ^{ - 1}}\left( {\cfrac{{3x + 1}}{2}} \right) + C$