Question

Find the integral of $f\left(x\right)=\frac{2x+2}{3x^2+2x+1}$

Answer 1

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

$\frac{2x+2}{3x^2+2x+1}=\frac{A\frac{d(3x^2+2x+1)}{dx}+B}{3x^2+2x+1}$

$\Rightarrow 2x+2=A(6x+2)+B$

On comparing the coefficient of constant and $x$ term, we get,
$6A=2$ and $2A+B=2$
$\Rightarrow A=\frac{1}{3}$ and $B=\frac{4}{3}$

$\Rightarrow \int f\left(x\right)=\frac{1}{3}\int \frac{6x+2}{3x^2+2x+1}dx+\frac{4}{3}\int \frac{dx}{3x^2+2x+1}$
$\Rightarrow \int f\left(x\right)=\frac{1}{3}{I}_1+\frac{4}{3}{I}_2+C$

$I_1=\int \frac{6x+2}{3x^2+2x+1}dx$
substitute, $3x^2+2x+1=t\Rightarrow (6x+2)dx=dt$
$\Rightarrow I_1=\int \frac{dt}{t}={\log }_e(3x^2+2x+1)$

$I_2=\int \frac{dx}{3x^2+2x+1}$
$=\frac{1}{3}\int \frac{dx}{x^2+2\cdot x\cdot \frac{1}{3}+{\left(\frac{1}{3}\right)}^2-\frac{1}{9}+\frac{1}{3}}$
$=\frac{1}{3}\int \frac{dx}{{(x+\frac{1}{3})}^2+{\left(\frac{\sqrt{2}}{3}\right)}^2}$
$=\frac{1}{3}\times \frac{3}{\sqrt{2}}{\tan }^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)$
$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)$

$\therefore f\left(x\right)=\frac{1}{3}{I}_1+\frac{4}{3}{I}_2+C$
$=\frac{1}{3}{\log }_e(3x^2+2x+1)+\frac{2\sqrt{2}}{3}{\tan }^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)+C$