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$\displaystyle\int{\dfrac{\left(\sqrt{x}+2\right)^2}{3\sqrt{x}}\;\mathrm{d}x} =\frac{2}{3}\big(\sqrt{x}+2\big) $ I do not know how to get $\dfrac{2}{3}$. I assume $\sqrt{x}+2 = u$, then $\mathrm{d}u= \dfrac{1}{2}\,x^{1/2}$

But it is wrong answer....

So help me

Thank you

Vlad
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user319316
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1 Answers1

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Assuming you mean $\displaystyle \int \frac{\left(\sqrt{x}+2\right)^2}{3\sqrt{x}}\,dx$:

Let $\,u=\sqrt{x}+2\,$ and $\,du=\dfrac{dx}{2\sqrt{x}}$

Thus, we now have $$\dfrac{2}{3} \int u^2 \,du=\dfrac{2u^3}{9}+C=\dfrac{2\left(\sqrt{x}+2\right)^3}{9}+C$$

Vlad
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zz20s
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  • No problem! You may want to consider accepting the answer by clicking the checkmark. Glad I could help! – zz20s Mar 02 '16 at 15:22