The correct option is B
43(x−1x+4)1/4+C
This integral can also be written as
I=∫dx((x−1)3(x+2)5]1/4⇒I=∫dx(x−1x+2)3/4(x+2)2
Now let,
(x−1x+2)=t⇒dt=3dx(x+2)2
Thus our integral becomes,
I=13∫1t3/4dt =13(t1/41/4)+C =43t1/4+C =43(x−1x+2)1/4+C
Thus, Option b. is correct.