Evaluate the integral $\int_0^{\ln(2)} \sqrt{(e^x-1)}dx$ 
Evaluate the integral $\int_0^{\ln(2)} \sqrt{(e^x-1)}dx$

Why is it wrong to...
$$\int_0^{\ln(2)} \sqrt{(e^x-1)} dx= \int_0^{\ln(2)} (e^x-1)^{1/2} dx= \frac{2}{3}(e^x-1)^{3/2} |_0^{\ln(2)}$$

You forgot to find the derivative of $e^x -1$ to replace $dx$ in the original integral. 
Put $e^x - 1 = u$. Then $e^x = u+1$.
Then $du = e^x\,dx\iff dx = \frac{du}{u+1}$
This gives us $$\int_0^1 \frac{u^{1/2}}{u+1} \,du$$
