Evaluate the integral by type1 or type 2 Evaluate $\displaystyle\int_{0}^{2} \int_{0}^{\log(x)}(x-1)\sqrt{1+e^y}\,dy\,dx$.
I have tried integration by substitution but can't connect to type 1 or type 2.
Any help. 
 A: I have no idea of what you mean by type-1 or type-2, but in any case:
$$\begin{eqnarray*}I=\int_{0}^{1}\int_{0}^{\log x}(x-1)\sqrt{1+e^{y}}\,dy\,dx &=& \int_{0}^{1}\int_{1}^{x}(x-1)\sqrt{1+z}\,\frac{dz}{z}\,dx\\&=&\int_{0}^{1}\int_{0}^{1-x}(1-x)\frac{\sqrt{2-z}}{1-z}\,dz\,dx\\&=&\int_{0}^{1}\int_{0}^{1-z}(1-x)\frac{\sqrt{2-z}}{1-z}\,dx\,dz\\&=&\int_{0}^{1}\frac{1-z^2}{2}\frac{\sqrt{2-z}}{1-z}\,dz\end{eqnarray*}$$
so:
$$ I = \frac{1}{2}\int_{0}^{1}(1+z)\sqrt{2-z}\,dx = \frac{1}{2}\int_{1}^{2}(3-z)\sqrt{z}\,dx=\color{red}{\frac{2}{5}\left(3\sqrt{2}-2\right)}.$$
A: Change the order of integration to obtain 
$$\begin{align}
I&\equiv\int_0^1\int_0^{\log x}(x-1)\sqrt{1+e^y}dydx\\\\&=-\int_{-\infty}^0\int_0^{e^y}(x-1)\sqrt{1+e^y}dxdy\\\\&=-\int_{-\infty}^0\left(\frac12 e^{2y}-e^y\right)\sqrt{1+e^y}dxdy
\end{align}$$
Then, letting $y\to \log y$, we have
$$\begin{align}
I&=-\int_0^1\left(\frac12 y-1\right)\sqrt{1+y}dy\\\\
&=-\int_1^2\left(\frac12 y-\frac32\right)y^{1/2}dy\\\\
&=\frac25\left(3\sqrt{2}-2\right)
\end{align}$$
