Integration by Parts Calculator

The integration by parts formula is ∫u dv = uv − ∫v du. It is derived from the product rule of differentiation and is used to integrate products of two functions by strategically splitting the integrand into u (to differentiate) and dv (to integrate).

A widely used guide is the LIATE rule: choose u in this priority order — Logarithmic, Inverse trigonometric, Algebraic (polynomials), Trigonometric, Exponential. The remaining part of the integrand becomes dv. For example, in ∫x·cos(x)dx, u = x (Algebraic) and dv = cos(x)dx (Trigonometric).

An integral is the reverse operation of differentiation. It calculates the accumulation of a quantity — geometrically, the area under a curve. Indefinite integrals give a family of antiderivatives (with a constant C), while definite integrals compute the exact net area between two bounds.

Use integration by parts when the integrand is a product of two different types of functions — such as a polynomial times an exponential (xe^x), a polynomial times a trigonometric function (x·sin x), or a logarithm times a polynomial (x·ln x). It is typically not needed for simple single-function integrals.

An indefinite integral produces a general antiderivative expressed as a function plus a constant C (e.g. sin(x) + C). A definite integral evaluates the antiderivative at an upper bound b and subtracts its value at a lower bound a, yielding a specific numerical result using the Fundamental Theorem of Calculus: F(b) − F(a).

Yes. When the remaining integral ∫v du is still a product of two function types, you apply integration by parts again. For example, ∫x²·eˣ dx requires two applications. In some cases like ∫eˣ·sin(x) dx, the formula is applied twice and then the original integral is solved algebraically.

LIATE stands for Logarithms, Inverse trig, Algebraic, Trigonometric, Exponential — it ranks function types by how easily they differentiate into simpler forms. Choosing u from earlier in the list tends to simplify the remaining integral ∫v du, making the overall computation more manageable.

Absolutely. For definite integrals ∫[a to b] u dv, the formula becomes [uv] evaluated from a to b, minus ∫[a to b] v du. You apply the bounds at the final step after all antiderivatives have been found, giving a precise numerical answer.