Integration of UV Formula — ∫u dv = uv − ∫v du

Integration of UV Formula — ∫u dv = uv − ∫v du

TL;DR

The integration of uv formula — integration by parts — is ∫u,dv=uv−∫v,du, the integration counterpart of the product rule for derivatives. This article derives it from the product rule, explains the ILATE rule for choosing u, works six examples in increasing difficulty (including the ∫ln⁡x,dx trick), and shows the mistakes that turn a one-line problem into an endless loop.

A Formula for Products That Refuse to Integrate

Differentiating a product is easy — the product rule handles xsin⁡x or xex in seconds. Integrating one is where students hit a wall, because there is no "product rule for integrals." The integration of uv formula is the closest thing: it doesn't integrate the product directly, it trades a hard integral for an easier one.

The integration of uv formula — also called integration by parts or the uv rule of integration — says:

∫u,dv = uv − ∫v,du

You split the integrand into a part u (which you'll differentiate) and a part dv (which you'll integrate), and the formula rewrites the original integral as a product minus a new integral — one you choose your split to make simpler.

What Is the Integration of UV Formula?

The integration of uv formula is the technique for integrating a product of two functions when no straightforward antiderivative exists. In its most-used form:

∫u,dv=uv−∫v,du

Here u and v are functions of x; you pick u and dv from the integrand, then compute du (by differentiating u) and v (by integrating dv). Written fully in terms of x:

∫u,w,dx=u∫w,dx−∫(u′∫w,dx)dx.

The point is never to integrate the product head-on. It is to swap the original integral for uv−∫v,du, where the new integral ∫v,du is simpler than the one you started with. Choose the split so that differentiating u shrinks it.

How Is the Integration of UV Formula Derived?

It comes straight from the product rule for derivatives. Start with two functions u and v:

ddx(uv) = u dv/dx + v du/dx.

Integrate both sides with respect to x. The left side integrates back to uv:

uv=∫u, dv/dx, dx + ∫v, du/dx, dx = ∫u, dv + ∫v, du.

Now rearrange to isolate the integral you want:

∫u,dv=uv−∫v,du.

That's the whole derivation. Integration by parts is the product rule run backwards and solved for one of its two integrals.

How Do You Choose u? The ILATE Rule

The formula only helps if the new integral ∫v,du is easier than the original. The choice of u decides that. The ILATE rule (also written LIATE) ranks function types by how good a choice they make for u — pick u from whichever type comes first:

Priority Type Example
I Inverse trigonometric arctan⁡x, arcsin⁡x
L Logarithmic ln⁡x, log⁡x
A Algebraic x, x², polynomials
T Trigonometric sin⁡x, cos⁡x
E Exponential eⁿ, 2ˡ

The logic: u should get simpler when differentiated, and dv should stay manageable when integrated. For ∫xln⁡x,dx, L beats A, so u=ln⁡x.

Examples of the Integration of UV Formula

Example 1

Evaluate ∫x,eˣ,dx.

By ILATE, Algebraic (A) beats Exponential (E), so u=x and dv=eˣ,dx. Then du=dx and v=eˣ:

∫x,eˣ,dx=uv−∫v,du=x,eˣ−∫eˣ,dx=x,eˣ−eˣ+C.

Final answer: ∫x,eˣ,dx=eˣ(x−1)+C.

Example 2

Evaluate ∫xcos⁡x,dx.

Let u=x (Algebraic) and dv=cos⁡x,dx. Then du=dx and v=sin⁡x:

∫xcos⁡x,dx=xsin⁡x−∫sin⁡x,dx=xsin⁡x+cos⁡x+C.

Final answer: ∫xcos⁡x,dx=xsin⁡x+cos⁡x+C.

Example 3

Evaluate ∫x²ln⁡x,dx.

By ILATE, Logarithmic (L) beats Algebraic (A), so u=ln⁡x and dv=x²,dx. Then du= rac{1}{x},dx and v= rac{x³}{3}:

∫x²ln⁡x,dx= rac{x³}{3}ln⁡x− rac{1}{3}∫x²,dx.

Final answer: ∫x²ln⁡x,dx= rac{x³}{3}ln⁡x− rac{x³}{9}+C.

Example 4

Evaluate ∫ln⁡x,dx.

Let u=ln⁡x and dv=1,dx. Then du= rac{1}{x},dx and v=x:

∫ln⁡x,dx=xln⁡x−x+C.

Final answer: ∫ln⁡x,dx=xln⁡x−x+C.

Example 5

Evaluate ∫arctan⁡x,dx.

Let u=arctan⁡x and dv=1,dx; then du= rac{1}{1+x²},dx and v=x:

∫arctan⁡x,dx=xarctan⁡x− rac{1}{2} ext{ln}(1+x²)+C.

Final answer: ∫arctan⁡x,dx=xarctan⁡x− rac{1}{2} ext{ln}(1+x²)+C.

Example 6

Evaluate ∫eˣsin⁡x,dx.

Let I=∫eˣsin⁡x,dx and let u=sin⁡x, dv=eˣ,dx:

I=eˣsin⁡x−∫eˣcos⁡x,dx.

Apply parts again to ∫eˣcos⁡x,dx with the same steps returning to I yields:

Final answer: ∫eˣsin⁡x,dx= rac{eˣ(sin⁡x−cos⁡x)}{2}+C.

Where the Integration of UV Formula Carries Real Weight

Tripping Points to Avoid

Mistake 1: Choosing u and dv backwards

Mistake 2: Dropping the minus sign before the second integral

Mistake 3: Forgetting the constant of integration

Conclusion

Practice These Before Moving On

  1. Evaluate ∫xsin⁡x,dx.
  2. Evaluate ∫x,ln⁡x,dx.
  3. Evaluate ∫eˣcos⁡x,dx using the loop method.