Mathematics

Solve :
$$\displaystyle\int { \cfrac { x }{ \sqrt { x+4 }  }  } dx$$


SOLUTION
$$\displaystyle\int{\dfrac{x}{\sqrt{x+4}}dx}$$

$$=\displaystyle\int{\dfrac{x+4-4}{\sqrt{x+4}}dx}$$

$$=\displaystyle\int{\dfrac{x+4}{\sqrt{x+4}}dx}-4\displaystyle\int{\dfrac{dx}{\sqrt{x+4}}}$$

$$=\displaystyle\int{{\left(x+4\right)}^{1-\frac{1}{2}}dx}-4\displaystyle\int{{\left(x+4\right)}^{-\frac{1}{2}}dx}$$

$$=\displaystyle\int{{\left(x+4\right)}^{\frac{1}{2}}dx}-4\displaystyle\int{{\left(x+4\right)}^{-\frac{1}{2}}dx}$$

We know that $$\displaystyle\int{{\left(ax+b\right)}^{n}}=\dfrac{1}{a\left(n+1\right)}{\left(ax+b\right)}^{n+1}$$

$$=\dfrac{{\left(x+4\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}-4\dfrac{{\left(x+4\right)}^{\frac{-1}{2}+1}}{\dfrac{-1}{2}+1}+c$$

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

$$=\dfrac{2}{3}{\left(x+4\right)}^{\frac{3}{2}}-8{\left(x+4\right)}^{\frac{1}{2}}+c$$

$$=\dfrac{2}{3}\left(x+4\right){\left(x+4\right)}^{\frac{1}{2}}-\dfrac{24}{3}{\left(x+4\right)}^{\frac{1}{2}}+c$$

$$=\dfrac{2}{3}{\left(x+4\right)}^{\frac{1}{2}}\left(x+4-12\right)+c$$

$$=\dfrac{2}{3}{\left(x+4\right)}^{\frac{1}{2}}\left(x-8\right)+c$$

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Subjective Medium Published on 17th 09, 2020
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