The correct option is
A (2x2−2)ln(x+1)−x2+2x4+c∫xlog(1+x)dx
=log(1+x)∫xdx−∫[ddxlog(1+x).∫x.dx]
=log(1+x)x22−∫11+x.x22dx
=x22log(x+1)−12∫x2−1+1(x+1)dx
=x22log(x+1)−12[∫(x−1)(x+1)(x+1)dx+∫1(x+1)dx]
=x22log(x+1)−12∫(x−1)dx−12log(x+1)
=12log(x+1)(x2−1)−12[x22−x]+c.
=12log(x+1)(x2−1)−12[x22−x]+c.
Hence, the answer is (2x2−2)ln(x+1)−x2+2x4+c