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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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