Question

Integrate:
$\int \frac{dx}{3x^2+13x-10}$

Answer 1

$\int \frac{dx}{3x^2+13x-10}$

$=\int \frac{dx}{3x^2+15x-2x-10}$

$=\int \frac{dx}{3x(x+5)-2(x+5)}$

$=\int \frac{dx}{(x+5)(3x-2)}$

Consider $\frac{1}{(x+5)(3x-2)}=\frac{A}{x+5}+\frac{B}{3x-2}$

$\Rightarrow 1=A(3x-2)+B(x+5)$

Put $x=-5$

$\Rightarrow 1=A(3\times -5-2)+0$

$\Rightarrow 1=A(-15-2)$

$\therefore A=\frac{-1}{17}$

Put $x=0$

$\Rightarrow 1=A(0-2)+B(0+5)$

$\Rightarrow -2A+5B=1$

$\Rightarrow -2\times \frac{-1}{17}+5B=1$

$\Rightarrow \frac{2}{17}+5B=1$

$\Rightarrow 5B=1-\frac{2}{17}=\frac{17-2}{17}=\frac{15}{17}$

$\therefore B=\frac{3}{17}$

$\therefore \int \frac{dx}{3x^2+13x-10}$

$=\frac{-1}{17}\int \frac{dx}{x+5}+\frac{3}{17}\int \frac{dx}{3x-2}$

$=\frac{-1}{17}\log (x+5)+\frac{3}{17}\times \frac{1}{3}\log (3x-2)$

$=\frac{-1}{17}\log (x+5)+\frac{1}{17}\log (3x-2)+c$

$=\frac{1}{17}\log \left(\frac{3x-2}{x+5}\right)+c$