TheoremProved

Integration by parts

Statement

If $u=u(x)$ and $v=v(x)$ are differentiable on an interval $K$ , then $\int u\,dv = uv - \int v\,du$ , i.e. $\int u(x)v'(x)\,dx = u(x)v(x) - \int u'(x)v(x)\,dx$ .

Why is it true?

Integration by parts reverses the product rule: it does not remove the integral, but it shifts the derivative from one factor ( $v$ ) onto the other ( $u$ ), which is exactly the right move when $u$ becomes simpler after differentiating (like $u=x$ becoming $u'=1$ ) while $v'$ stays easy to antidifferentiate — turning a hard product like $\int x e^x\,dx$ into an easy one.

Proof sketch

By the product rule, $(uv)' = u'v + uv'$ . Rearranging, $uv' = (uv)' - u'v$ .

Take the antiderivative (with respect to $x$ , on $K$ ) of both sides. On the right, $(uv)'$ obviously has antiderivative $uv$ itself (up to a constant, by the uniqueness theorem), so $\int uv'\,dx = \int (uv)'\,dx - \int u'v\,dx = uv - \int u'v\,dx$ .

Writing $dv = v'(x)\,dx$ and $du=u'(x)\,dx$ turns this into the compact mnemonic form $\int u\,dv = uv - \int v\,du$ . In practice, one factor is chosen as $u$ (to be differentiated) and the rest as $dv$ (to be antidifferentiated), typically preferring to differentiate the factor that simplifies (polynomial, logarithm) and antidifferentiate the factor that stays manageable ( $e^x$ , $\sin x$ , $\cos x$ ).

Topics that use this theorem

Antiderivative

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

James Stewart (2015). Calculus: Early Transcendentals
Michael Spivak (2008). Calculus
Manuel Bronstein (1998). Symbolic Integration Tutorial