How do you integrate #int lnx/x^7# by integration by parts method? Calculus Techniques of Integration Integration by Parts 1 Answer Konstantinos Michailidis Aug 12, 2016 We have that #int lnx*[-x^-6/6]'dx=-1/6*lnx*x^-6+1/6*int(lnx)'*x^-6dx= -1/6*lnx*x^-6+1/6*int 1/x*x^-6dx= -1/6*lnx*x^-6+1/6*int x^-7dx= -1/6*lnx*x^-6+1/6*int (x^-6/6)'dx= -1/6*lnx*x^-6-(x^-6/36)+c= -(6*lnx+1)/(36*x^6)+c# 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 2729 views around the world You can reuse this answer Creative Commons License