Mathematics · Ch 1 — Integrals

Integral of the Type

1.6.1

Integral of the Type

7.6.1 Integrals of the Form $\int e^x [f(x) + f'(x)] \, dx$

This section develops a shortcut for integrating expressions where $e^x$ multiplies the sum of a function and its derivative. It emerges from integration by parts, but once understood it lets you write down the answer almost immediately.

Derivation of the Formula

I = \int e^x [f(x) + f'(x)] \, dx

Split the integrand into two integrals:

I = \int e^x f(x) \, dx + \int e^x f'(x) \, dx = I_1 + \int e^x f'(x) \, dx \quad \text{where} \quad I_1 = \int e^x f(x) \, dx \qquad(1)

Evaluate $I_1$ by parts, taking $f(x)$ as the first function and $e^x$ as the second function:

I_1 = e^x f(x) - \int e^x f'(x) \, dx + C

Substitute this back into equation (1):

I = \left[ e^x f(x) - \int e^x f'(x) \, dx + C \right] + \int e^x f'(x) \, dx

The two integrals $\int e^x f'(x) \, dx$ cancel each other exactly, leaving:

\boxed{\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C}

Important

The formula works only when the integrand is exactly $e^x$ multiplied by the sum of a function and its own derivative. If the derivative term is missing or incorrect, this shortcut does not apply.

How to Use the Formula

Recognise when a given integrand can be written as $e^x [f(x) + f'(x)]$ . Look for:

A factor of $e^x$ (or a constant multiple of it)
A remaining factor that is the sum of some function $f(x)$ and its derivative $f'(x)$

Once you identify $f(x)$ , the answer is simply $e^x f(x) + C$ . To verify an answer, differentiate $e^x f(x)$ — you should get back $e^x [f(x) + f'(x)]$ . …

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Integral of the Type