I haven't see in your calculations to multiply by π .
I found after integration from 0 to 1/2 V = 5.0477...
check it and do not forget to multiply by π.
$\displaystyle \begin{align*} V &= \pi \int_0^{\frac{1}{2}}{ \left( 1 + \frac{1}{4x^2 + 1} \right) ^2 \, dx } \\ &= \pi \int_0^{\frac{\pi}{4}}{ \left\{ 1 + \frac{1}{4 \left[ \frac{1}{2}\tan{(\theta)} \right] ^2 + 1 } \right\} ^2 \, \frac{1}{2}\sec^2{(\theta)}\,d\theta } \textrm{ after making the substitution } 2x = \tan{(\theta)} \\ &= \frac{\pi}{2}\int_0^{\frac{\pi}{4}}{ \left[ 1 + \frac{1}{\tan^2{(\theta)} + 1 } \right] ^2 \sec^2{(\theta)}\,d\theta } \\ &= \frac{\pi}{2}\int_0^{\frac{\pi}{4}}{ \left[ 1 + \frac{1}{\sec^2{(\theta)} } \right] ^2 \sec^2{(\theta)}\,d\theta} \\ &= \frac{\pi}{2} \int_0^{\frac{\pi}{4}}{ \left[ 1 + \cos^2{(\theta)} \right] ^2 \sec^2{(\theta)} \,d\theta } \\ &= \frac{\pi}{2} \int_0^{\frac{\pi}{4}}{ \left[ 1 + 2\cos^2{(\theta)} + \cos^4{(\theta)} \right] \sec^2{(\theta)}\,d\theta } \\ &= \frac{\pi}{2} \int_0^{\frac{\pi}{4}}{ \sec^2{(\theta)} + 2 + \cos^2{(\theta)} \,d\theta } \\ &= \int_0^{\frac{\pi}{4}}{ \sec^2{(\theta)} + 2 + \frac{1}{2} + \frac{1}{2}\cos{(2\theta)} \,d\theta } \\ &= \frac{\pi}{2} \int_0^{\frac{\pi}{4}}{ \sec^2{(\theta)} + \frac{5}{2} + \frac{1}{2}\cos{(2\theta)} \,d\theta } \\ &= \frac{\pi}{2} \left[ \tan{(\theta)} + \frac{5}{2}\theta + \frac{1}{4}\sin{(2\theta)} \right]_0^{\frac{\pi}{4}} \\ &= \frac{\pi}{2} \left\{ \left[ \tan{ \left( \frac{\pi}{4} \right) } + \frac{5}{2}\left( \frac{\pi}{4} \right) + \frac{1}{4}\sin{ \left( \frac{\pi}{2} \right) } \right] - \left[ \tan{(0)} + \frac{5}{2} \left( 0 \right) + \frac{1}{4}\sin{(0)} \right] \right\} \\ &= \frac{\pi}{2} \left[ \left( 1 + \frac{5\pi}{8} + \frac{1}{4} \right) - \left( 0 + 0 + 0 \right) \right] \\ &= \frac{\pi}{2} \left( \frac{5}{4} + \frac{5\pi}{8} \right) \\ &= \frac{5\pi}{8} \left( 1 + \frac{\pi}{2} \right) \\ &= \frac{5\pi}{16} \left( 2 + \pi \right) \end{align*}$