Mathematics

$\displaystyle \int \dfrac{x}{\sqrt{x+2}}\ dx$

SOLUTION

We have,

$\int{\dfrac{x}{\sqrt{x+2}}dx}$

Let $x+2=t$

$x=t-2$

$\dfrac{dx}{dt}=1$

$dx=dt$

Therefore,

$\int{\dfrac{t-2}{\sqrt{t}}dt}$

$\Rightarrow \int{\left( \dfrac{t}{\sqrt{t}}-\dfrac{2}{\sqrt{t}} \right)dt}$

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

$\Rightarrow \dfrac{{{t}^{\frac{1}{2}+1}}}{\dfrac{1}{2}+1}-2\dfrac{{{t}^{\frac{-1}{2}+1}}}{\dfrac{-1}{2}+1}+c$

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

$\Rightarrow \dfrac{2}{3}{{t}^{\frac{3}{2}}}-4{{t}^{\frac{1}{2}}}+c$

Put $t=x+2$

$\Rightarrow \dfrac{2}{3}{{\left( x+2 \right)}^{\frac{3}{2}}}-4{{\left( x+2 \right)}^{\frac{1}{2}}}+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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