Solids of Revolution Question

**The Problem**

The finite region bounded by the curve $\displaystyle y\ =\ e^x\ +\ 1$, the y-axis, and the line $\displaystyle x\ =\ ln\ 2$, is rotated through 360 degrees about the x-axis. Show that the volume of the solid formed is $\displaystyle \frac{\Pi}{2}(7\ +\ ln\ 4)$.

**Attempt**

Volume = $\displaystyle Integral(\Pi\ y^2\ dx)$, with limits $\displaystyle 0\ and\ ln\ 2.$

This becomes: Volume = $\displaystyle \Pi\ Integral((e^x\ +\ 1)^2)$, with limits $\displaystyle 0\ and\ ln\ 2.$

(We can take pi outside the integral and multiply the result by it later).

Integrating term by term:

$\displaystyle Integral((e^x\ +\ 1)^2)\ =\ \frac{e^{2x}}{2}\ +\ 2e^x + 1$.

Substituting $\displaystyle ln\ 2$:

$\displaystyle \frac{e^{2\ ln\ 2}}{2}\ +\ 2e^{ln\ 2}\ +\ 1\ =\ 7.$

Substituting $\displaystyle 0$:

$\displaystyle \frac{e^{2(0)}}{2}\ +\ 2e^{0}\ +\ 1\ =\ 3\ \frac{1}{2}.$

Subtracting:

$\displaystyle 7\ -\ 3\ \frac{1}{2}\ =\ 3\ \frac{1}{2}$.

So my answer is:

$\displaystyle 3\ \frac{1}{2}\ \Pi$.

So I don't know how they've reached the answer given.

Any hints?