Question

Evaluate: $\int \frac{5x-2}{1+2x+3x^2}dx$

Answer 1

${\int }^{}\frac{5x-2}{3x^2+2x+1}dx$

putting $5x-2=A\frac{d}{dx}\left(3x^2+2x+1\right)$

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

$\Rightarrow 5x-2=6Ax+2A+B$

on comparing both sides we get

$6A=5$$\Rightarrow A=\frac{5}{6}$

and $2A+B=-2$

$\Rightarrow \frac{10}{6}+B=-2$

$\Rightarrow B=\frac{-11}{3}$

Now, ${\int }^{}\frac{5x-2}{3x^2+2x+1}dx={\int }^{}\frac{\frac{5}{6}\left(6x+2\right)-\frac{11}{3}}{3x^2+2x+1}dx$

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

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

$=\frac{5}{6}{\int }^{}\frac{dt}{t}-\frac{11}{9}{\int }^{}\frac{dx}{x^2+\frac{2}{3}x+\frac{1}{9}-\frac{1}{9}+\frac{1}{3}}$

$=\frac{5}{6}\log \left|t\right|-\frac{11}{9}{\int }^{}\frac{dx}{{\left(x+\frac{1}{3}\right)}^2+{\left(\frac{\sqrt{2}}{3}\right)}^2}$

$=\frac{5}{6}\log \left|3x^2+2x+1\right|-\frac{11}{3\sqrt{2}}{\tan }^{-1}\left(\frac{x+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\right)+c$ (since ${\int }^{}\frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+c$)

$=\frac{5}{6}\log \left|3x^2+2x+1\right|-\frac{11}{3\sqrt{2}}{\tan }^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)+c$