Mathematics

# $\int { \left( x+2 \right) \sqrt { 3x+5 } } dx$

##### SOLUTION

We have,

$\int{\left( x+2 \right)\sqrt{3x+5}}dx$

Let

$3x+5=t\Rightarrow x=\dfrac{t-5}{3}$

$3dx=dt$

$dx=\dfrac{dt}{3}$

So,

$\int{\left( \dfrac{t-5}{3}+2 \right)}\sqrt{t}dx$

$\Rightarrow \int{\left( \dfrac{t-5+6}{3} \right)}\sqrt{t}dx$

$\Rightarrow \int{\left( \dfrac{t+1}{3} \right)\sqrt{t}dt}$

$\Rightarrow \dfrac{1}{3}\int{\left( t+1 \right)\sqrt{t}}dt$

$\Rightarrow \dfrac{1}{3}\int{t\sqrt{t}}dt+\dfrac{1}{3}\int{\sqrt{t}}dt$

$\Rightarrow \dfrac{1}{3}\int{{{t}^{\frac{3}{2}}}dt}+\dfrac{1}{3}\int{{{t}^{\frac{1}{2}}}dt}$

On integrating and we get,

$\Rightarrow \dfrac{1}{3}\dfrac{{{t}^{\frac{3}{2}+1}}}{\dfrac{3}{2}+1}+\dfrac{1}{3}\dfrac{{{t}^{\frac{1}{2}+1}}}{\dfrac{1}{2}+1}+C$

$\Rightarrow \dfrac{1}{3}\dfrac{{{t}^{\frac{5}{2}}}}{\dfrac{5}{2}}+\dfrac{1}{3}\dfrac{{{t}^{\dfrac{}{2}}}}{\dfrac{3}{2}}+C$

$\Rightarrow \dfrac{2{{t}^{\frac{5}{2}}}}{15}+\dfrac{2{{t}^{\frac{3}{2}}}}{9}+C$

now, put

$t=3x+5$

Then,

$\Rightarrow \dfrac{2{{\left( 3x+5 \right)}^{\frac{5}{2}}}}{15}+\dfrac{2{{\left( 3x+5 \right)}^{\frac{3}{2}}}}{9}$

Hence, this is the answer.

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