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 $$t=x+2$$

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


Hence, this is the answer.
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