If ∫dxx3(x−1)1/2=x−1(3x+2)4x2+Ktan−1⁡x−1+C then the value of K is

Put x−1=t2 so that dx=2t dt and ∫dxx3(x−1)1/2=∫2tdtt2+13t=2∫dtt2+13Now we use formula of ∫dyy2+k2n to write∫dtt2+13=t4t2+12+34∫dtt2+12 and ∫dtt2+12=t2t2+1+12∫dtt2+1=t2t2+1+12tan−1⁡tHence ∫dxx3(x−1)1/2                   =t2t2+12+3t4t2+1+34tan−1⁡t+C=14t2+122t+3t3+3t+34tan−1⁡t+C=x−1(3x+2)4x2+34tan−1⁡x−1+C.