∫1(x−1)3(x+2)51/4dx is equal to
43x−1x+21/4+C
43x+2x−11/4+C
13x−1x+21/4+C
13x+2x−11/4+C
∫1(x−1)3(x+2)51/4dx=∫1x−1x+23/4(x+2)2dx=13∫1t3/4dt put x−1x+2=t⇒3(x+2)2dx=dt=13t1/414+C=43t1/4+C=43x−1x+21/4+C