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Integration by Parts Calculator

Last updated: March 15, 2026

Calculator

Results

Definite integral

3.718282

u·v at lower bound

1

u·v at upper bound

8.154845

∫v·du at lower helper

2

∫v·du at upper helper

5.436564

Upper minus lower

1

Safe c used in calculation

1

Results

Definite integral

3.718282

u·v at lower bound

1

u·v at upper bound

8.154845

∫v·du at lower helper

2

∫v·du at upper helper

5.436564

Upper minus lower

1

Safe c used in calculation

1

The Integration by Parts Calculator evaluates definite integrals of the form $$\int_a^b u \, dv$$ where $$u = ax + b$$ is a linear function and $$dv$$ is one of $$e^{cx}dx$$, $$\sin(cx)dx$$, or $$\cos(cx)dx$$. This technique, one of the most important methods in integral calculus, transforms a difficult integral into a simpler one by strategically choosing which part of the integrand to differentiate and which to integrate.

Integration by parts is derived from the product rule for differentiation. If $$u$$ and $$v$$ are differentiable functions of $$x$$, then the product rule states $$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$$. Integrating both sides and rearranging yields the integration by parts formula: $$\int u \, dv = uv - \int v \, du$$. This identity was first formulated by Brook Taylor in 1715, building on the foundational work of Newton and Leibniz.

The key insight in applying integration by parts is the choice of $$u$$ and $$dv$$. The LIATE rule provides a practical heuristic: choose $$u$$ to be the function that appears earliest in the list Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. For the integrands handled by this calculator, the linear function $$ax + b$$ is algebraic while the other factor is trigonometric or exponential, so the linear function is always chosen as $$u$$. This ensures that $$du = a \, dx$$ is simpler than $$u$$, and the remaining integral $$\int v \, du$$ can be evaluated directly.

When $$dv = e^{cx}dx$$, the antiderivative is $$v = \frac{1}{c}e^{cx}$$, and the correction integral $$\int v \, du = \frac{a}{c}\int e^{cx}dx = \frac{a}{c^2}e^{cx}$$. When $$dv = \sin(cx)dx$$, we get $$v = -\frac{1}{c}\cos(cx)$$, and $$\int v \, du = -\frac{a}{c^2}\sin(cx)$$. When $$dv = \cos(cx)dx$$, we get $$v = \frac{1}{c}\sin(cx)$$, and $$\int v \, du = \frac{a}{c^2}\cos(cx)$$. In each case, the correction integral has a known closed form, so the full result is obtained without further iteration.

Integration by parts appears throughout science and engineering. In physics, it is used to derive the equations of motion from Lagrangian mechanics, to evaluate Fourier transforms, and to compute expectation values in quantum mechanics. In electrical engineering, Laplace transforms of products rely on integration by parts. In probability theory, moments of continuous distributions often require this technique. In signal processing, the convolution of signals is closely related to iterated integration by parts.

This calculator evaluates the definite integral over a specified interval $$[a, b]$$. It displays the boundary term $$[uv]_a^b$$ and the correction term $$\left[\frac{a}{c} \cdot V(x)\right]_a^b$$ separately, allowing you to see exactly how the integration by parts formula produces the final result. Enter your parameters and bounds to get an instant, step-transparent computation.

Visual Analysis

How It Works

Integration by parts applies the formula:

$$\int_a^b u \, dv = \left[uv\right]_a^b - \int_a^b v \, du$$

With $$u = ax + b$$ (so $$du = a\,dx$$) and the chosen $$dv$$, the antiderivative $$v$$ depends on the function type:

For $$dv = e^{cx}dx$$:

$$v = \frac{e^{cx}}{c}, \quad \int v\,du = \frac{a}{c^2}e^{cx}$$

$$\text{Result} = \left[\frac{(ax+b)e^{cx}}{c}\right]_a^b - \left[\frac{a \cdot e^{cx}}{c^2}\right]_a^b$$

For $$dv = \sin(cx)dx$$:

$$v = -\frac{\cos(cx)}{c}, \quad \int v\,du = -\frac{a}{c^2}\sin(cx)$$

$$\text{Result} = \left[-\frac{(ax+b)\cos(cx)}{c}\right]_a^b - \left[-\frac{a\sin(cx)}{c^2}\right]_a^b$$

For $$dv = \cos(cx)dx$$:

$$v = \frac{\sin(cx)}{c}, \quad \int v\,du = \frac{a}{c^2}\cos(cx)$$

$$\text{Result} = \left[\frac{(ax+b)\sin(cx)}{c}\right]_a^b - \left[\frac{a\cos(cx)}{c^2}\right]_a^b$$

The calculator evaluates each bracket at the upper and lower bounds and computes the difference to produce the definite integral value.

Understanding Your Results

The Definite Integral Result is the net signed area under the curve $$u \cdot \frac{dv}{dx}$$ between the specified bounds. A positive value indicates the function lies predominantly above the x-axis; a negative value means it lies predominantly below.

The u*v at Upper/Lower Bound values show the boundary term $$[uv]_a^b$$ evaluated at each endpoint. The difference of these gives the first part of the integration by parts formula.

The Correction values represent $$\frac{a}{c} \cdot V(x)$$ where $$V(x)$$ is the antiderivative of $$v$$. The correction integral $$\int v\,du$$ is subtracted from the boundary term. When $$a = 0$$ (constant $$u$$), the correction vanishes and integration by parts reduces to straightforward antidifferentiation of $$dv$$.

Worked Examples

Integral of (2x+1)e^x from 0 to 1

Inputs

a2
b1
c1
dvTypeexp
lower0
upper1

Results

definiteResult3.436564
uvUpper8.154845
uvLower1
correctionUpper5.436564
correctionLower2

u = 2x+1, dv = e^x dx, v = e^x. At x=1: uv = 3*e = 8.1548, correction = 2e = 5.4366. At x=0: uv = 1*1 = 1, correction = 2*1 = 2. Result = (8.1548 - 1) - (5.4366 - 2) = 7.1548 - 3.4366 = 3.7183. Verifiable: integral of (2x+1)e^x = (2x-1)e^x, evaluated [0,1] = e - (-1) = e+1 = 3.7183.

Integral of (x+2)sin(2x) from 0 to pi

Inputs

a1
b2
c2
dvTypesin
lower0
upper3.14159

Results

definiteResult-2.570796
uvUpper2.570796
uvLower-1
correctionUpper-0.000001
correctionLower0

u = x+2, dv = sin(2x)dx, v = -cos(2x)/2. At x=pi: uv = (pi+2)*(-cos(2pi)/2) = -(pi+2)/2 = -2.5708. At x=0: uv = 2*(-1/2) = -1. Correction uses integral of -cos(2x)/2, which is -sin(2x)/4. Evaluated at bounds both give approximately 0. Result = (-2.5708-(-1)) - (0-0) = -1.5708.

Frequently Asked Questions

Integration by parts is a technique that transforms the integral of a product of two functions into a (hopefully simpler) integral. It is based on the product rule for differentiation: $$\int u\,dv = uv - \int v\,du$$. The method works best when differentiating $$u$$ simplifies it and integrating $$dv$$ is straightforward.

The LIATE heuristic (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) recommends choosing as $$u$$ the function type that appears earliest in this list. A linear polynomial is algebraic, which comes before trigonometric and exponential. Differentiating a linear function produces a constant, which simplifies the remaining integral.

When $$a = 0$$, $$u = b$$ is a constant. Integration by parts reduces to $$b \int dv$$, which is simply $$b \cdot v$$. The correction integral vanishes because $$du = 0$$. Effectively, this becomes direct integration of the $$dv$$ function scaled by the constant $$b$$.

Yes. When $$u$$ is a polynomial of degree $$n$$, integration by parts must be applied $$n$$ times (or use the tabular method) to fully resolve the integral. Each application reduces the polynomial degree by one. This calculator handles the case $$n = 1$$ (linear $$u$$), which requires exactly one application.

The tabular method (also called the DI method) organizes repeated integration by parts into a table with alternating signs. One column lists successive derivatives of $$u$$, the other lists successive antiderivatives of $$dv$$. Products are taken diagonally with alternating signs. This shortcut avoids writing out each iteration explicitly.

The parameter $$c$$ appears in the denominator of the antiderivative, so $$c = 0$$ would cause division by zero. The calculator guards against this by replacing very small values of $$c$$ with a tiny nonzero value (0.0001). For practical use, ensure $$c \neq 0$$.

Sources & Methodology

Stewart, James. Calculus: Early Transcendentals, 9th ed., Cengage, 2020. Thomas, George B., et al. Thomas' Calculus, 15th ed., Pearson, 2023. Apostol, Tom M. Calculus, Vol. 1, 2nd ed., Wiley, 1967. Weisstein, Eric W., "Integration by Parts," MathWorld.
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