Mathematics

# Solve :$\displaystyle \int { \cfrac { 2x+1 }{ \sqrt { 3x+2 } } } dx\quad$

##### SOLUTION
$\displaystyle\int{\dfrac{2x+1}{\sqrt{3x+2}}dx}$

$=\displaystyle\int{\dfrac{2\left(x+1\right)-1}{\sqrt{3x+2}}dx}$

$=2\displaystyle\int{\dfrac{\left(x+1\right)}{\sqrt{3x+2}}dx}-2\displaystyle\int{\dfrac{dx}{\sqrt{3x+2}}}$

$=\dfrac{2}{3}\displaystyle\int{\dfrac{\left(3x+2+1\right)}{\sqrt{3x+2}}dx}-2\displaystyle\int{\dfrac{dx}{\sqrt{3x+2}}}$

$=\dfrac{2}{3}\displaystyle\int{\dfrac{\left(3x+2\right)}{\sqrt{3x+2}}dx}+\dfrac{2}{3}\displaystyle\int{\dfrac{dx}{\sqrt{3x+2}}}-2\displaystyle\int{\dfrac{dx}{\sqrt{3x+2}}}$

$=\dfrac{2}{3}\displaystyle\int{\dfrac{\left(3x+2\right)}{\sqrt{3x+2}}dx}-\dfrac{4}{3}\displaystyle\int{\dfrac{dx}{\sqrt{3x+2}}}$

$=\dfrac{2}{3}\displaystyle\int{{\left(3x+2\right)}^{1-\frac{1}{2}}dx}-\dfrac{4}{3}\displaystyle\int{{\left(3x+2\right)}^{-\frac{1}{2}}dx}$

$=\dfrac{2}{3}\displaystyle\int{{\left(3x+2\right)}^{\frac{1}{2}}dx}-\dfrac{4}{3}\displaystyle\int{{\left(3x+2\right)}^{-\frac{1}{2}}dx}$

$=\dfrac{2}{3}\times\dfrac{1}{3}\dfrac{{\left(3x+2\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}-\dfrac{4}{3}\times\dfrac{1}{3}\dfrac{{\left(3x+2\right)}^{\frac{-1}{2}+1}}{\dfrac{-1}{2}+1}+c$

$=\dfrac{2}{9}\dfrac{{\left(3x+2\right)}^{\frac{3}{2}}}{\dfrac{3}{2}}-\dfrac{8}{9}{\left(3x+2\right)}^{\frac{1}{2}}+c$

$=\dfrac{4}{27}{\left(3x+2\right)}^{\frac{3}{2}}-\dfrac{8}{9}{\left(3x+2\right)}^{\frac{1}{2}}+c$

$=\dfrac{4}{27}\left(3x+2\right){\left(3x+2\right)}^{\frac{1}{2}}-\dfrac{8}{9}{\left(3x+2\right)}^{\frac{1}{2}}+c$

$=\left[\dfrac{4}{27}\left(3x+2\right)-\dfrac{24}{27}\right]{\left(3x+2\right)}^{\frac{1}{2}}+c$

$=\dfrac{1}{27}\left(12x+8-24\right){\left(3x+2\right)}^{\frac{1}{2}}+c$

$=\dfrac{4}{27}\left(3x-4\right){\left(3x+2\right)}^{\frac{1}{2}}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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