How do you integrate #(x^4)(lnx)#? Calculus Techniques of Integration Integration by Parts 1 Answer Shwetank Mauria Jul 26, 2016 #intlnx xxx^4dx=x^5/5(lnx-1/5)# Explanation: WE can use integration by parts #intudv=uv-intvdu# Let #u=lnx# and #v=x^5/5# Hence #du=dx/x# and #dv=x^4dx# and #intudv=uv=intvdu# is #intlnx xxx^4dx=intudv=uv-intvdu# = #x^5/5xxlnx-intx^5/5xxdx/x# = #(lnx xx x^5)/5-1/5intx^4dx# = #(lnx xx x^5)/5-x^5/25# = #x^5/5(lnx-1/5)# 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 13554 views around the world You can reuse this answer Creative Commons License