∫(x−x3)1/3x4 dx
38 (1x2−1)4/3+C
−38 (1x2−1)4/3+C
38 (1+1x2)4/3+C
−38 (1+1x2)4/3+C
∫x[1x2−1]13x4dx= −12∫t13dt
1x2−1=t⇒−2x3dx=dt
=−12t43(43)+C
=−38(1x2−1)43+C