Integrate (lnx)/x^2 dx between limits 1 and 5

Let I = integral[(lnx)/x^2 dx] for simplicity.Firstly, we realise we must use integration by parts. This is:Integral [u(x)v'(x) dx] = u(x)v(x) - Integral[u'(x)v(x) dx]So we can see that, by letting u(x)=lnx and v'(x)=1/x^2, we have:I = (lnx)(-1/x) - integral[(1/x)(-1/x) dx] = -(lnx)/x + integral [1/x^2 dx] = -(lnx)/x - 1/x (+C would be used for indefinite integral; where there are no limits)Plug in the limits, we have:-ln5/5-1/5+ln1/1+1/1=4/5 - (ln5)/5or (4-ln5)/5

Answered by Jimmy L. Maths tutor

3222 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How to integrate by parts


How do I differentiate a function of x and y with respect to x?


A curve has the equation: x^3 - x - y^3 - 20 = 0. Find dy/dx in terms of x and y.


Derive the quadratic formula. From it, write down the determinant and explain, how is it related to the roots of a quadratic equation.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences