Help Integrate: sqrt(2+x^(1/2))

Hello, speedengine!

Quote:

$\int \sqrt{2 + \sqrt{x}}\:dx$

Answer: . $\frac{4}{15}(\sqrt{x}+2)^{\frac{3}{2}}(3\sqrt{x} - 4) + C$

With "nested" radicals, it often works if we let $u$ equal something inside.

Let: $2 + \sqrt{x}\:=\:u\quad\Rightarrow\quad \sqrt{x}\:=\:u-2\quad\Rightarrow\quad x$ $\:=\:(u - 2)^2\quad\Rightarrow\quad dx \:=\:2(u-2)\,du$

Substitute: . $\int u^{\frac{1}{2}}\cdot2(u - 2)\,du\;=\;2\int\left(u^{\frac{3}{2}} - 2u^{\frac{1}{2}}\right)\,du$ $\;=\;2\left(\frac{2}{5}u^{\frac{5}{2}} - \frac{4}{3}u^{\frac{3}{2}}\right) + C$

Factor out $\frac{2}{15}:\;\;\frac{4}{15}\left(3u^{\frac{5}{2} } - 10u^{\frac{3}{2}}\right) + C$

Factor out $u^{\frac{3}{2}}:\;\;\frac{4}{15}u^{\frac{3}{2}}(3u - 10) + C$

Since $u\;=\;\sqrt{x} + 2$ , we have: . $\frac{4}{15}(\sqrt{x} + 2)^{\frac{3}{2}}\left[3(\sqrt{x} + 2) - 10\right] + C$

. . . $= \;\frac{4}{15}(\sqrt{x} + 2)^{\frac{3}{2}}(3\sqrt{x} + 6 - 10) + C\;=$ $\;\boxed{\frac{4}{15}(\sqrt{x} + 2)^{\frac{3}{2}}(3\sqrt{x} - 4) + C}$