Mathematics

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


SOLUTION
$$\displaystyle\int{\dfrac{dx}{3{x}^{2}+13x-10}}$$
$$=\displaystyle\int{\dfrac{dx}{3{x}^{2}+15x-2x-10}}$$
$$=\displaystyle\int{\dfrac{dx}{3x\left(x+5\right)-2\left(x+5\right)}}$$
$$=\displaystyle\int{\dfrac{dx}{\left(x+5\right)\left(3x-2\right)}}$$
Consider $$\dfrac{1}{\left(x+5\right)\left(3x-2\right)}=\dfrac{A}{x+5}+\dfrac{B}{3x-2}$$
$$\Rightarrow 1=A\left(3x-2\right)+B\left(x+5\right)$$
Put $$x=-5$$
$$\Rightarrow 1=A\left(3\times -5-2\right)+0$$
$$\Rightarrow 1=A\left(-15-2\right)$$
$$\therefore A=\dfrac{-1}{17}$$
Put $$x=0$$
$$\Rightarrow 1=A\left(0-2\right)+B\left(0+5\right)$$
$$\Rightarrow -2A+5B=1$$
$$\Rightarrow -2\times \dfrac{-1}{17}+5B=1$$
$$\Rightarrow \dfrac{2}{17}+5B=1$$
$$\Rightarrow 5B=1-\dfrac{2}{17}=\dfrac{17-2}{17}=\dfrac{15}{17}$$
$$\therefore B=\dfrac{3}{17}$$
$$\therefore \displaystyle\int{\dfrac{dx}{3{x}^{2}+13x-10}}$$
$$=\dfrac{-1}{17}\displaystyle\int{\dfrac{dx}{x+5}}+\dfrac{3}{17}\displaystyle\int{\dfrac{dx}{3x-2}}$$
$$=\dfrac{-1}{17}\log{\left(x+5\right)}+\dfrac{3}{17}\times \dfrac{1}{3}\log{\left(3x-2\right)}$$
$$=\dfrac{-1}{17}\log{\left(x+5\right)}+\dfrac{1}{17}\log{\left(3x-2\right)}+c$$
$$=\dfrac{1}{17}\log{\left(\dfrac{3x-2}{x+5}\right)}+c$$
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Subjective Hard Published on 17th 09, 2020
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