Integration by Parts: Choose u and dv, Then Check the Minus Sign

Integration by parts uses ∫u dv = uv − ∫v du. Choose u so its derivative simplifies the remaining work, and choose dv so you can find an antiderivative v.

The product rule explains the formula

Differentiating uv gives u′v + uv′. Integrating and rearranging yields the parts formula. The minus sign is therefore structural, not an optional convention. Integration of a product does not generally equal the product of two antiderivatives.

A common useful choice is to differentiate a polynomial factor and integrate an exponential factor. Each differentiation lowers the polynomial degree. This is a strategy rather than a universal rule; always inspect whether the new integral is easier than the original.

Worked example

Find ∫x e^(2x) dx. Choose u = x, so du = dx. Choose dv = e^(2x)dx, so v = e^(2x)/2. Substituting into the formula gives x e^(2x)/2 − ∫e^(2x)/2 dx.

Integrate the remaining exponential to obtain x e^(2x)/2 − e^(2x)/4 + C. Equivalently, write e^(2x)(x/2 − 1/4) + C. The extra factor 1/2 appears twice: once when finding v and again when integrating the remaining exponential.

Check by differentiation. The derivative is e^(2x)(x − 1/2) + e^(2x)/2 = x e^(2x). The two non-x terms cancel, returning the original integrand exactly.

A logarithm can be the u factor

For ∫ln x dx on x > 0, regard the integrand as ln x × 1. Choose u = ln x and dv = dx. Then du = dx/x and v = x, giving x ln x − ∫1 dx = x ln x − x + C.

This example shows why the absence of a visibly written product does not rule out parts. In contrast, when an inner expression’s derivative already appears as another factor, substitution may be shorter. Decide based on the structure rather than applying one preferred technique to every integral.

Definite integrals

With fixed limits, use [uv] at both endpoints and keep the same limits on the remaining integral. Do not evaluate only the upper endpoint. A final number has no arbitrary +C, because the constant cancels in endpoint subtraction.

Common mistake

List all four quantities u, dv, du and v before substituting. Confusing dv with v, or dropping the minus sign before the second integral, changes the answer. Finish with a derivative check whenever the final antiderivative is simple enough.

u, dv, du, v alag likho; formula ka minus sign sambhal kar.

Three practice questions with worked answers

1. Evaluate ∫x e^x dx.

   Take u = x and dv = e^x dx. Then the result is x e^x − ∫e^x dx = (x − 1)e^x + C.

2. Evaluate ∫ln x dx for x > 0.

   Choose u = ln x and dv = dx. The parts formula gives x ln x − x + C; differentiating gives ln x + 1 − 1.

3. Evaluate ∫ from 0 to 1 of x e^x dx.

   Use (x − 1)e^x at the endpoints. The value at 1 is 0, and at 0 it is −1. The definite integral is 0 − (−1) = 1.

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References and authorship

Original examples and practice questions, prepared with AI assistance and editorial arithmetic review. Related definitions and methods: OpenStax: Integration by Parts. No textbook or exam-board affiliation is claimed.