What is #int (ln(x+1)/(x^2)) dx#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Monzur R. Apr 12, 2018 #intln(x+1)/x^2dx=ln(x/(x+1))-ln(x+1)/x+"c"# Explanation: For #intln(x+1)/x^2dx#, we integrate by parts Let #u=ln(x+1)rArrdu=1/(x+1)dx# and #dv=1/x^2dxrArrv=-1/x# So #intln(x+1)/x^2dx=-ln(x+1)/x+int1/(x(x+1))dx=int1/x-1/(x+1)dx-ln(x+1)/x=lnx-ln(x+1)-ln(x+1)/x# Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 1759 views around the world You can reuse this answer Creative Commons License