Mathematics

Evaluate $\displaystyle\int { \dfrac { x }{ 1+\sqrt { x } } } dx$.

SOLUTION
Let, I  $=\displaystyle\int{\dfrac{x}{1+\sqrt{x}}dx}$

Put $x={t}^{2}\Rightarrow dx=2tdt$

$I=\displaystyle\int{\dfrac{{t}^{2}}{1+t}2t \, dt}$

$=2\displaystyle\int{\dfrac{{t}^{3}}{1+t}dt}$

$=2\displaystyle\int{\dfrac{{t}^{3}+1-1}{1+t}dt}$

$=2\displaystyle\int{\dfrac{{t}^{3}+1}{1+t}dt}-2\displaystyle\int{\dfrac{1}{1+t}dt}$

$=2\displaystyle\int{\dfrac{\left(t+1\right)\left({t}^{2}-t+1\right)}{1+t}dt}-2\displaystyle\int{\dfrac{1}{1+t}dt}$

$=2\displaystyle\int{\left({t}^{2}-t+1\right)dt}-2\displaystyle\int{\dfrac{1}{1+t}dt}$

$=2\left[\dfrac{{t}^{3}}{3}-\dfrac{{t}^{2}}{2}+t\right]-2\log{\left|1+t\right|}+c$

$=\dfrac{2{t}^{3}}{3}-\dfrac{2{t}^{2}}{2}+2t-2\log{\left|1+t\right|}+c$

$=\dfrac{2{t}^{3}}{3}-{t}^{2}+2t-2\log{\left|1+t\right|}+c$

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

$=\dfrac{2x\sqrt{x}}{3}-x+2\sqrt{x}-2\log{\left|1+\sqrt{x}\right|}+c$

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Subjective Medium Published on 17th 09, 2020
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