Question

Integrate the function $\frac{5x-2}{1+2x+3x^2}$

Answer 1

Let $5x-2=A\frac{d}{dx}(1+2x+3x^2)+B$
$\Rightarrow 5x-2=A(2+6x)+B$
Equating the coefficient of $x$ and constant term on both sides, we obtain
$5=6A\Rightarrow A=\frac{5}{6}$
and $2A+B=-2\Rightarrow B=-\frac{11}{3}$
$\therefore 5x-2=\frac{5}{6}(2+6x)+(-\frac{11}{3})$
$\Rightarrow \int \frac{5x-2}{1+2x+3x^2}dx=\int \frac{\frac{5}{6}(2+6x)-\frac{11}{3}}{1+2x+3x^2}dx$
Let $I_1=\int \frac{2+6x}{1+2x+3x^2}dx$ and $I_2=\int \frac{1}{1+2x+3x^2}dx$
$\therefore \int \frac{5x-2}{1+2x+3x^2}dx=\frac{5}{6}{I}_1-\frac{11}{3}{I}_2$ ............. (1)
$I_1=\int \frac{2+6x}{1+2x+3x^2}dx$
Let $1+2x+3x^2=t$
$\Rightarrow (2+6x)dx=dt$
$\therefore I_1=\int \frac{dt}{t}$
$=\log |t|=\log |1+2x+3x^2|$ .....(2)
$I_2=\int \frac{1}{1+2x+3x^2}dx$
$1+2x+3x^2+$ can be written as $1+3(x^2+\frac{2}{3}x)$.
Therefore,
$1+3(x^2+\frac{2}{3}x)$
$=1+3(x^2+\frac{2}{3}x+\frac{1}{9}-\frac{1}{9})$
$=1+3{(x+\frac{1}{3})}^2-\frac{1}{3}$
$=\frac{2}{3}+3{(x+\frac{1}{3})}^2$
$=3[{(x+\frac{1}{3})}^2+\frac{2}{9}]$
$=3[{(x+\frac{1}{3})}^2+{\left(\frac{\sqrt{2}}{3}\right)}^2]$
$I_2=\frac{1}{3}\int \frac{1}{[{(x+\frac{1}{3})}^2+{\left(\frac{\sqrt{2}}{3}\right)}^2]}dx$
$=\frac{1}{3}\left[\frac{1}{\frac{\sqrt{2}}{3}}{\tan }^{-1}\left(\frac{x+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\right)\right]$
$=\frac{1}{3}\left[\frac{3}{\sqrt{2}}{\tan }^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)\right]$
$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)$ ..........(3)
Substituting equations (2) and (3) in equation (1), we obtain
$\int \frac{5x-2}{1+2x+3x^2}dx=\frac{5}{6}[\log |1+2x+3x^2|]-\frac{11}{3}\left[\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)\right]+C$
$=\frac{5}{6}\log |1+2x+3x^2|-\frac{11}{3\sqrt{2}}{\tan }^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)+C$