The value of ∫dx(x−1)3(x+2)54 is
3(x−1)1/4−(5/3)(x+2)3/4+C
4(x−14)−5(x+2)54+C
43x−1x+2+C4
none of these
I=∫dx4(x−1)3(x+2)5=∫x−1x+2−3/41(x+2)2dx
Now put x−1x+2=t to obtain
I=13∫t−3/4dt=43t1/4+C