U Substitution

[Starblazer]

I am trying to solve
\(\displaystyle \int {\frac{{{{(\sqrt x + 2)}^2}}}{{5\sqrt x }}} dx\)
using u substitution.

Do I make \(\displaystyle u = \sqrt x \) or \(\displaystyle u = \sqrt x + 2\) or is it done another way?

Thank You

 

[Starblazer]

I managed to solve it

\(\displaystyle \begin{array}{l}
\int {\frac{{{{(\sqrt x + 2)}^2}}}{{5\sqrt x }}} dx\\
u = \sqrt x + 2\\
du = \frac{1}{2}{x^{ - \frac{1}{2}}}\\
2du = {x^{^{ - \frac{1}{2}}}}\\
\int {\frac{{{u^2}}}{5}} 2du\\
= \frac{1}{5}2\int {{u^2}du} \\
= \frac{2}{5}\frac{{{u^3}}}{3} + C\\
= \frac{2}{{15}}{\left( {\sqrt x + 2} \right)^3} + C
\end{array}\)

 

[mmm4444bot]

\(\displaystyle du = \frac{1}{2}{x^{ - \frac{1}{2}}}\)\(\displaystyle dx\)

\(\displaystyle 2du = {x^{^{ - \frac{1}{2}}}}\)\(\displaystyle dx\)

\(\displaystyle dx = \dfrac{2du}{\sqrt{x}}\)

Rest looks okay, to me. :cool:

 

[lookagain]

I managed to solve it


\(\displaystyle \int {\dfrac{{{{(\sqrt x + 2)}^2}}}{{5\sqrt x }}} dx\)


\(\displaystyle u \ = \ \sqrt x + 2 \)


\(\displaystyle du \ = \ \dfrac{1}{2}{x^{ - \frac{1}{2}}}dx\)


\(\displaystyle 2du \ = \ {x^{^{ - \frac{1}{2}}}}dx\)


\(\displaystyle \int {\bigg(\dfrac{{{u^2}}}{5}\bigg)} (2)du\)


= \(\displaystyle \ \bigg(\dfrac{1}{5}\bigg)(2)\int {{u^2}du} \)


= \(\displaystyle \ \dfrac{2}{5}\bigg(\dfrac{u^3}{3}\bigg) + C\)


= \(\displaystyle \ \dfrac{2}{{15}}{\left( {\sqrt x + 2} \right)^3} + C\)

Starblazer, I made some changes. Not counting the "dx" symbols that mmm4444bot amended,
you need appropriate grouping symbols and/or multiplication symbols to show certain products.