I am trying to solve a question that involves integration by substitution.
Let $\displaystyle I = \int^{1}_{0}\frac{\sqrt{x}}{2- \sqrt{x}}dx$
Using the substitution $u = 2-\sqrt{x}$ show that $\displaystyle I = \int^{2}_{1}\frac{2(2-u)^2}{u}dx$
I started by rearranging the equation for $u$, $\sqrt{x} = 2 - u$. The replacing instances of $\sqrt{x}$ in the equation with the substitution. $\displaystyle I = \int^{u}_{l} \frac{2-u}{u}dx$, however I'm not really sure where to go from here.
The second component of the question is as follows.
Hence show that $I = 8 * ln(2) - 5$
How would you go about solving both components of this question?
If $u=2-\sqrt x$, then $\mathrm du=-\frac1{2\sqrt x}\,\mathrm dx$. So\begin{align}\int_0^1\frac{\sqrt x}{2-\sqrt x}\,\mathrm dx&=-2\int_0^1\frac x{2-\sqrt x}\times\frac1{2\sqrt x}\,\mathrm dx\\&=-2\int_2^1\frac{(2-u)^2}u\,\mathrm du\\&=\int_1^2\frac{2(2-u)^2}u\,\mathrm du.\end{align}In order to compute this integral, note that$$\frac{2(2-u)^2}u=\frac8u-8+2u.$$