∫1(√32v)2+34√32
$∫1(√32v)2+34√32$
I assume that this is meant to be
√32∫1(√32v)2+34dv=√32∫134v2+34dv=√32∫134(v2+1)dv=√3×42×3∫1(v2+1)dv=2√3∫11+v2dv=2√3tan−1v+c
I think that is correct.