Integration by Parts Applier

Math Calculus • Integrals

Written by STEM Calculators Team Published December 28, 2025 Updated February 24, 2026

4. Integration By Parts Applier

Implements \(\int u\,dv = uv - \int v\,du\) with LIATE suggestions, repeated application / tabular view, optional definite bounds, and a shaded-area graph in definite mode.

Inputs

Integrand \(f(x)\)

Supported: + − * / ^, parentheses, variable x, constants pi, e, sin cos tan, ln log, sqrt, abs, exp. Implicit multiplication allowed: 2x, (x+1)(x-1), 2sin(x).

Mode

Definite mode enables shaded area on \([a,b]\).

Choose \(u\)

If the integrand is a product, you can force a specific \(u\).

Output format

Plot constant \(C\) for \(F(x)+C\)

Numeric tolerance

Presets

Click to auto-fill and compute.

Options

Show steps Tabular view (when applicable) Show grid Shade (definite mode) Show \(f(x)\) Show \(F(x)+C\) Show \(F'(x)\)

Ready

Graph

Drag to pan • wheel/pinch to zoom • \(f(x)\), \(F(x)+C\), \(F'(x)\). In definite mode, shaded region shows signed area on \([a,b]\).

Center \(c\)

Half-width \(w\)

Legend

\(f(x)\) \(F(x)+C\) \(F'(x)\)

x: 0, y: 0, zoom(px/unit): 60

Result

Enter \(f(x)\) and click Evaluate.

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4. Integration By Parts Applier

A compact reference for Integration by Parts, LIATE choice, tabular method, and definite integrals.

Core formula

\[ \int u\,dv = u\,v - \int v\,du \]

This comes from the product rule: \(\frac{d}{dx}(uv) = u'v + uv'\). Identify \(u' = \frac{du}{dx}\) and \(v' = \frac{dv}{dx}\), then rearrange and integrate.

How to choose \(u\): LIATE

A common heuristic is to choose \(u\) so that \(du\) becomes simpler.

\[ \text{LIATE: } \textbf{L}ogarithmic \;>\; \textbf{I}nverse trig \;>\; \textbf{A}lgebraic \;>\; \textbf{T}rig \;>\; \textbf{E}xponential \]

Logarithmic: \(\ln(x)\), \(\log(x)\)
Algebraic: polynomials like \(x^2\), \(3x-1\)
Trig: \(\sin(x)\), \(\cos(x)\), \(\tan(x)\)
Exponential: \(e^x\), \(e^{ax+b}\)

Worked example

\[ \int x^2 e^x\,dx \]

Choose \(u=x^2\) (algebraic), \(dv=e^x\,dx\) (exponential). Then \(du=2x\,dx\) and \(v=e^x\).

\[ \int x^2 e^x\,dx = x^2 e^x - \int 2x e^x\,dx \]

Apply IBP again on \(\int 2x e^x\,dx\) to finish:

\[ \int x^2 e^x\,dx = x^2 e^x - 2x e^x + 2e^x + C \]

Tabular method (repeated IBP)

When \(u\) is a polynomial and \(dv\) is \(e^{ax}\), \(\sin(ax)\), or \(\cos(ax)\), repeated IBP is systematic. Differentiate the polynomial until zero; integrate the trig/exp repeatedly; combine with alternating signs.

\[ \int P(x)\,e^{ax}\,dx \quad\text{or}\quad \int P(x)\sin(ax)\,dx \quad\text{or}\quad \int P(x)\cos(ax)\,dx \]

Definite integrals with IBP

For definite integrals, the same formula applies, but you evaluate the boundary term:

\[ \int_a^b u\,dv = \Big[u\,v\Big]_a^b - \int_a^b v\,du \]

On the graph, the shaded region on \([a,b]\) represents the signed area under \(f(x)\): regions below the \(x\)-axis subtract.

Frequently Asked Questions

What is the integration by parts formula used by this calculator?

It uses integral u dv = u v - integral v du. The identity comes from rearranging the product rule d(uv)/dx = u'v + uv'.

How does LIATE help choose u in integration by parts?

LIATE is a heuristic ordering to pick u so that du becomes simpler: Logarithmic, Inverse trig, Algebraic, Trig, Exponential. Choosing u earlier in this list often reduces the remaining integral faster.

When is the tabular method useful for integration by parts?

Tabular integration is useful when repeated integration by parts is needed, especially for a polynomial times an exponential or trig function. Differentiating the polynomial to zero and repeatedly integrating the other factor produces a structured shortcut.

How does integration by parts work for definite integrals?

The boundary term must be evaluated at the limits: integral from a to b u dv = [u v] from a to b - integral from a to b v du. The calculator uses the same identity but applies the bounds to the u v term.

What does the shaded region mean in definite mode?

The shaded region on [a,b] represents the signed area under f(x). Areas above the x-axis add to the integral value and areas below the x-axis subtract.

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Frequently Asked Questions

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