Mathematics

Integrate with respect to $$x$$:
$$\dfrac {x}{\sqrt {x+2}}$$


SOLUTION

Consider the given integral.

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

 

Put,

$$u=x+2\Rightarrow du=dx$$

 

Therefore,

$$ I=\int{\dfrac{u-2}{\sqrt{u}}du} $$

$$ I=\int{\left( \sqrt{u}-\dfrac{2}{\sqrt{u}} \right)du} $$

$$ I=\int{\sqrt{u}du}-2\int{\dfrac{1}{\sqrt{u}}du} $$

$$ I=\dfrac{2}{3}{{\left( u \right)}^{3/2}}-4\sqrt{u}+C $$

 

Now, return the value of $$x$$.

$$ I=\dfrac{2}{3}{{\left( x+2 \right)}^{3/2}}-4\sqrt{x+2}+C $$

$$ I=\dfrac{2\sqrt{x+2}\left( x+2-6 \right)}{3}+C $$

$$ I=\dfrac{2\sqrt{x+2}\left( x-4 \right)}{3}+C $$

 

Hence, this is the required value of the integral.
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