$\cfrac{{5x - 2}}{{1 + 2x + 3{x^2}}}$

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

We put , $5x - 2 = A\left( {\cfrac{d}{{dx}}\left( {3{x^2} + 2x + 1} \right)} \right) + B$

$\Rightarrow$ $5x - 2 = A\left( {6x + 2} \right) + B$

….(i)
Comparing coefficients of x in (i),

we get
$5 = 6A$ $\Rightarrow$ $A = \cfrac{5}{6}$

Comparing the constant terms in (i),

we get
$- 2 = 2A + B$
$\Rightarrow$ $- 2 = 2 \times \cfrac{5}{6} + B$ $\Rightarrow$ $- 2 = \cfrac{5}{3} + B$ $\Rightarrow$ $B = - \cfrac{{11}}{3}$

$\therefore$ $I = \int {\cfrac{{\cfrac{5}{6}\left( {6x + 2} \right) - \cfrac{{11}}{3}}}{{3{x^2} + 2x + 1}}dx}$

$\Rightarrow$ $I = \cfrac{5}{6}\int {\cfrac{{6x + 2}}{{3{x^2} + 2x + 1}}} dx - \cfrac{{11}}{3}\int {\cfrac{{dx}}{{3{x^2} + 2x + 1}}}$

$\Rightarrow$ $I = \cfrac{5}{6}{I_1} - \cfrac{{11}}{3}{I_2}$

…..(ii)
Let ${I_1} = \int {\cfrac{{6x + 2}}{{3{x^2} + 2x + 1}}} dx$

Put $3{x^2} + 2x + 1 = t$ $\Rightarrow$ $\left( {6x + 2} \right)dx = dt$

$\therefore$ ${I_1} = \int {\cfrac{{dt}}{t} = \log \left| t \right| + {C_1} = \log \left| {3{x^2} + 2x + 1} \right|} + {C_1}$

….(iii)
Let ${I_2} = \int {\cfrac{{dx}}{{3{x^2} + 2x + 1}}} = \cfrac{1}{3}\int {\cfrac{{dx}}{{{x^2} + \cfrac{2}{3}x + \cfrac{1}{3}}}}$

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

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

$= \cfrac{1}{3} \times \cfrac{1}{{\sqrt 2 /3}}{\tan ^{ - 1}}\left( {\cfrac{{x + \cfrac{1}{3}}}{{\cfrac{{\sqrt 2 }}{3}}}} \right) = \cfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\cfrac{{3x + 1}}{{\sqrt 2 }}} \right) + {C_2}$

….(iv)
From (ii), (iii) and (iv)

we get
$I = \cfrac{5}{6}\log \left( {3{x^2} + 2x + 1} \right) - \cfrac{{11}}{{3\sqrt 2 }}{\tan ^{ - 1}}\left( {\cfrac{{3x + 1}}{{\sqrt 2 }} + C} \right)$

$\left[ {C = \cfrac{5}{6}{C_1} - \cfrac{{11}}{3}{C_2}} \right]$

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