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

JL
Answered by Jimmy L. Maths tutor

3300 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

When using the trapezium rule to approximate area underneath a curve between 2 limits, what is the effect of increasing the number of strips used?


A ball is projected at an angle b from the horizontal. With initial velocity V the ball leaves the ground at point O and hits the ground at point A. If Vcos(b) = 6u and Vsin(b) = 2.5u, how long does the ball take to travel between O and A.


Find dy/dx when y = 5x^6 + 4x*sin(x^2)


By forming and solving a quadratic equation, solve the equation 5*cosec(x) + cosec^2(x) = 2 - cot^2(x) in the interval 0<x<2*pi, giving the values of x in radians to three significant figures.


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