Reverse the product rule strategically

Try

Differentiate x e^x. The result is e^x+x e^x, so rearranging suggests a way to integrate x e^x: subtract the simpler exponential contribution from the product. This is the origin of integration by parts, not an arbitrary recipe for choosing letters.

Predict

For x²e^x, predict whether differentiating the polynomial or integrating it makes the next problem simpler. Repeated differentiation lowers its degree; integrating it raises the degree. A useful choice of parts makes progress rather than merely producing a different-looking integral.

Experiment

Choose u=x and dv=e^x dx. Then du=dx and v=e^x.

Step	Expression
Product term uv	x e^x
Remaining integral ∫v du	∫e^x dx
Result	x e^x−e^x+C
Derivative check	e^x+x e^x−e^x=x e^x

For a definite version from 0 to 1, evaluate the primitive at both endpoints: zero minus negative one gives 1. Draw a positive integrand on this interval and use its sign to check that a negative computed result would be impossible.

Understand

The product derivative (uv)′=u′v+uv′ integrates to [uv]ₐᵇ=∫ₐᵇu′v+∫ₐᵇuv′ under sufficient regularity, such as continuously differentiable u and v. Rearranging gives ∫u dv=uv−∫v du, with endpoint evaluation on the product term for a definite integral. The notation dv means a chosen derivative expression whose antiderivative v you can obtain.

A factor need not be visibly multiplied by another nonconstant function. For ∫ln x dx, choose u=ln x and dv=dx on x>0. Then du=dx/x and v=x, giving x ln x−∫1 dx=x ln x−x+C. Evaluating from 1 to e gives 1. This example illustrates why differentiating a logarithm can simplify a problem even though the original integrand is not written as a product.

Repeated parts handles x²e^x. First obtain x²e^x−2∫x e^x dx. Apply the previous result to get e^x(x²−2x+2)+C. Differentiate the final expression: the polynomial and its derivative combine to x². This verification catches sign errors that can accumulate across repeated steps.

Sometimes the original integral returns. Let I=∫e^x cos x dx. Parts gives I=e^x cos x+∫e^x sin x dx. A second application gives the remaining integral as e^x sin x−I. Therefore 2I=e^x(cos x+sin x), so a primitive is half that expression. Treat the calculation as an identity between antiderivatives up to constants, then verify the proposed primitive directly to avoid confusion about arbitrary constants.

Parts is not always the best choice. For x(2x+1), expansion gives a simple polynomial. For a clear inner derivative, substitution may be shorter. Choosing a method means recognizing structure and comparing the complexity of what remains, not obeying a rigid mnemonic without checking its result.

Apply

In the practice bank, name u, dv, du and v, and retain the minus sign before the remaining integral. For definite problems, include both endpoint terms. In the cyclic example, solve the resulting equation instead of repeating the same steps indefinitely.

Continue

Next decompose rational functions into simpler pieces. Polynomial division and factor structure determine the appropriate form before any integration takes place.