∫1x1−x3dx is equal to

We have,l=∫1x1−x3dx⇒ I=−13∫1x31−x3−3x2dx⇒ I=−13∫1x31−x3d1−x3⇒ I=−13∫11−t2t22tdt, where t2=1−x3⇒ t=−23∫11−t2dt=23∫1t2−12dt=13log⁡t−1|t+1|+C⇒ I=13log⁡1−x3−11−x3+1+C