Mathematics

Evaluate the following integral:
$$\int { \cfrac { { \left( 1+\sqrt { x }  \right)  }^{ 2 } }{ \sqrt { x }  }  } dx$$


SOLUTION
Let 
$$t=1+\sqrt{x}\Rightarrow\,dt=\dfrac{1}{2\sqrt{x}}dx$$

$$\Rightarrow\,2dt=\dfrac{1}{\sqrt{x}}dx$$

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

$$=2\displaystyle\int{{t}^{2}\,dt}$$

$$=\dfrac{2{t}^{3}}{3}+c$$

$$=\dfrac{2{\left(1+\sqrt{x}\right)}^{3}}{3}+c$$ where $$t=1+\sqrt{x}$$

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Subjective Medium Published on 17th 09, 2020
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