Question

Integrate the following functions.
$\int \frac{5x-2}{1+2x+3x^2}dx.$

Answer 1

$\int \frac{5x-2}{1+2x+3x^2}dx$
Let $5x-2=A\frac{d}{dx}(1+2x+3x^2)+B$
$\Rightarrow 5x-2=A(2+6x)+B\Rightarrow 5x-2=6Ax+(2A+B)$
On equating the coefficient of x and constant on both sides, we get
$5=6A\Rightarrow A=\frac{5}{6}\text{and}2A+B=-2\Rightarrow B=-\frac{11}{3}\therefore 5x-2=\frac{5}{6}(2+6x)+(-\frac{11}{3})\therefore I=\int \frac{5x-2}{1+2x+3x^2}dx=\int \frac{\frac{5}{6}(2+6x)-\frac{11}{3}}{1+2x+3x^2}dx$
Let $I_1=\int \frac{2+6x}{1+2x+3x^2}dx\text{and I\_2}=\int \frac{1}{1+2x+3x^2}dx$
$\therefore I=\frac{5}{6}{I}_1-\frac{11}{3}{I}_2$

Now, $I_1=\int \frac{2+6x}{1+2x+3x^2}dx$
Let $1+2x+3x^2=t$
$\Rightarrow (2+6x)dx=dt\therefore I_1=\int \frac{dt}{t}=log|t|+C_1\Rightarrow I_1=log|1+2x+3x^2|+C_1.......\left(ii\right)$
Also, $I_2=\int \frac{1}{1+2x+3x^2}dx$

$1+2x+3x^2\text{can be written as}1+3(x^2+\frac{2}{3}x)\text{Therefore},1+3(x^2+\frac{2}{3}x)=1+3(x^2+\frac{2}{3}x+\frac{1}{9}-\frac{1}{9})=1+3{(x+\frac{1}{3})}^2-\frac{1}{3}=\frac{2}{3}+3{(x+\frac{1}{3})}^2=3\left[\right(x+\frac{1}{3}{)}^2+\frac{2}{9}]=3\left[\right(x+\frac{1}{3}{)}^2+\left(\frac{\sqrt{2}}{3}{)}^2\right]\therefore I_2=\frac{1}{3}\int \frac{dx}{\left[\right(x+\frac{1}{3}{)}^2+\left(\frac{\sqrt{2}}{3}{)}^2\right]}=\frac{1}{3}\left[\frac{1}{\frac{\sqrt{2}}{3}}ta{n}^{-1}\right(\frac{x+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\left)\right]+C_2[\because \int \frac{dx}{a^2+x^2}=\frac{1}{a}ta{n}^{-1}(\frac{x}{a}\left)\right]=\frac{1}{3}\left[\frac{3}{\sqrt{2}}ta{n}^{-1}\right(\frac{3x+1}{\sqrt{2}}\left)\right]+C_2=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)+C_2.....\left(iii\right)$

On substituting the values of $I_1$ and $I_2$ from Eqs. (ii)and (iii)in Eq. (i), we get

$I=\frac{5}{6}[log|1+2x+3x^2|]-\frac{11}{3}\left[\frac{1}{\sqrt{2}}ta{n}^{-1}\right(\frac{3x+1}{\sqrt{2}}\left)\right]+C(\because \frac{5}{6}{C}_1-\frac{11}{3}{C}_2=C)=\frac{5}{6}log|1+2x+3x^2|-\frac{11}{3\sqrt{2}}ta{n}^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)+C$