What is the value of $\int_0^{\log 2}\int_{-1}^1ye^{xy} \,dx\,dy\,$?

Question

The question is to integrate the following $$\int_0^{\log 2}\int_{-1}^1ye^{xy} \,dx\,dy$$

I have solved but i am getting a different answer.the original answer is $1/2$.

Answer 1

In the second line of your proof, your lower limit of integration changes from $0$ to $2$. If you instead have

$$\begin{align} \int_0^{\ln 2}e^y+e^{-y} \ dy&=\int_0^{\ln 2}e^y \ dy + \int_0^{\ln 2} -e^{-y} \ dy \\ &=(2-1)+\left( \frac{1}{2}-1\right)=\frac{1}{2} \end{align}$$