How do you find the integral of #x^2 / sqrt (x-1) dx#? Calculus Techniques of Integration Integration by Parts 1 Answer Ratnaker Mehta Jun 2, 2018 # 2/15*sqrt(x-1)(3x^2+4x+8)+C.# Explanation: Let, #I=intx^2/sqrt(x-1)dx#. Subst. #x-1=t^2. :. x=t^2+1. :. dx=2tdt#. #:. I=int(t^2+1)^2/t*2tdt#, #=2int(t^4+2t^2+1)dt#, #=2(t^5/5+2*t^3/3+t)#, #=(2t)/15(3t^4+10t^2+15)#, #=2/15*sqrt(x-1){3(x-1)^2+10(x-1)+15}#. # rArr I=2/15*sqrt(x-1)(3x^2+4x+8)+C.# Enjoy Maths.! Answer link Related questions How do I find the integral #int(x*ln(x))dx# ? How do I find the integral #int(cos(x)/e^x)dx# ? How do I find the integral #int(x*cos(5x))dx# ? How do I find the integral #int(x*e^-x)dx# ? How do I find the integral #int(x^2*sin(pix))dx# ? How do I find the integral #intln(2x+1)dx# ? How do I find the integral #intsin^-1(x)dx# ? How do I find the integral #intarctan(4x)dx# ? How do I find the integral #intx^5*ln(x)dx# ? How do I find the integral #intx*2^xdx# ? See all questions in Integration by Parts Impact of this question 62225 views around the world You can reuse this answer Creative Commons License