Consider the given integral.
I=∫10x3x2+1dx
I=∫10(x−xx2+1)dx
I=∫10(x−2x2(x2+1))dx
I=[x22]10−12[loge(x2+1)]10
I=[122−0]−12[loge(12+1)−loge(0+1)]
I=12−12[loge(2)−loge(1)]
I=12−12[loge(2)−0]
I=12−12loge2
Hence, this is the answer.