Why does integration by parts work?

Let's consider the form of the formula we use for integration by parts: the integral of u * dv/dx = uv - the integral of v * du/dx. We know that integration is the inverse of differentiation, so we should be able to differentiate both sides to get back to u * dv/dx. The left hand side of this equation obviously satisfies this. d/dx(uv - integral of v * du/dx) = d/dx(uv) - d/dx(integral of v * du/dx).
For d/dx(uv), we'll use the product rule: d/dx(uv) = du/dx * v + u * dv/dx. For d/dx(integral of v * du/dx), we'll just use the fact that integration is the inverse of differentiation, d/dx(integral of v * du/dx) = v * du/dx. So, d/dx(uv) - d/dx(integral of v * du/dx) = (du/dx * v + u * dv/dx) - (du/dx * v) = u * dv/dx. Since the derivative of uv - integral of v * du/dx = u * dv/dx, the integral of u * dv/dx = uv - integral of v * du/dx.

TS
Answered by Tomas S. Maths tutor

4598 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Solve the equation |3x + 4| = |3x - 11|


The equation 2x^2 + 2kx + (k + 2) = 0, where k is a constant, has two distinct real roots. Show that k satisfies k^2 – 2k – 4 > 0


Differentiation basics: What is it?


If x=-2,1,2 and the y intercept is y=-8 for y=ax^3+bx^2+cx+d, what is a, b, c and d


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning