Hi,

Originally Posted by

**Beard** a)Integral of: 1/sqrtx(2+sqrtx).dx = ln 16, between the values of 36 and 0 and where the substitution is x=u^2

$\displaystyle I=\int_0^{36}\frac{1}{\sqrt{x}\left(2+\sqrt{x}\rig ht)}\,\mathrm{d}x$

Let $\displaystyle x=u^2$. We have $\displaystyle \frac{\mathrm{d}x}{\mathrm{d}u}=2u$ hence $\displaystyle \mathrm{d}x=2u\,\mathrm{d}u$.

$\displaystyle I=\int_{\sqrt{0}}^{\sqrt{36}} \frac{1}{\sqrt{u^2}\left(2+\sqrt{u^2}\right)}\cdot 2u\,\mathrm{d}u$

Since $\displaystyle u\geq0,\, \sqrt{u^2}=|u|=u$ so

$\displaystyle

\begin{aligned}

I&=2\int_0^6 \frac{1}{u\left(2+u\right)}\cdot u\,\mathrm{d}u\\

&=2\int_0^6 \frac{1}{2+u}\,\mathrm{d}u\\

\end{aligned}$

Can you take it from here ?