Integral of e^x[f(x) + f'(x)]

The shortcut formula

Whenever an integrand has the exact shape $e^x$ times $[f(x) + f'(x)]$ , it integrates instantly, no substitution or parts needed:

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

This follows directly from the product rule run in reverse: $\frac{d}{dx}\left[e^xf(x)\right] = e^xf(x) + e^xf'(x) = e^x[f(x)+f'(x)]$ .

Why it works

Differentiate $e^xf(x)$ with the product rule:

\dfrac{d}{dx}\left[e^xf(x)\right] = e^x\cdot f(x) + e^x\cdot f'(x) = e^x\left[f(x) + f'(x)\right]

So $e^xf(x)$ is an antiderivative of $e^x[f(x) + f'(x)]$ , and integrating just reverses that step. The whole skill is spotting $f$ : find a term in the bracket whose derivative is the other term.

Spotting f(x) in some common brackets

Bracket	$f(x)$	Integral
$\sin x + \cos x$	$\sin x$	$e^x\sin x + C$
$\tan x + \sec^2 x$	$\tan x$	$e^x\tan x + C$
$\ln x + \dfrac{1}{x}$	$\ln x$	$e^x\ln x + C$
$x^2 + 2x$	$x^2$	$x^2e^x + C$
$\dfrac{1}{x} - \dfrac{1}{x^2}$	$\dfrac{1}{x}$	$\dfrac{e^x}{x} + C$

Worked example

Find $\displaystyle\int e^x\left(\dfrac{1}{x} - \dfrac{1}{x^2}\right)dx$ .

Solution: Let $f(x) = \dfrac{1}{x}$ . Then $f'(x) = -\dfrac{1}{x^2}$ , exactly matching the second term. So the integrand is $e^x[f(x)+f'(x)]$ , and:

\int e^x\left(\dfrac{1}{x} - \dfrac{1}{x^2}\right)dx = e^xf(x) + C = \dfrac{e^x}{x} + C

Worked example: f has to be uncovered

Find $\displaystyle\int e^x\dfrac{x + 1}{(x + 2)^2}\,dx$ .

Solution: The bracket doesn't look like $f + f'$ yet. Write the numerator as $(x + 2) - 1$ and split:

\dfrac{x + 1}{(x + 2)^2} = \dfrac{x + 2}{(x + 2)^2} - \dfrac{1}{(x + 2)^2} = \dfrac{1}{x + 2} - \dfrac{1}{(x + 2)^2}

With $f(x) = \dfrac{1}{x + 2}$ , $f'(x) = -\dfrac{1}{(x + 2)^2}$ , which is exactly the second term:

\int e^x\left[\dfrac{1}{x + 2} - \dfrac{1}{(x + 2)^2}\right]dx = \dfrac{e^x}{x + 2} + C

Worked example: a definite integral

Evaluate $\displaystyle\int_0^1 e^x(x+1)\,dx$ .

Solution: With $f(x) = x$ , $f'(x) = 1$ , so the integrand is $e^x[f(x)+f'(x)]$ and the antiderivative is $xe^x$ :

\int_0^1 e^x(x+1)\,dx = \left[xe^x\right]_0^1 = e - 0 = e \approx 2.718

Common mistakes

Not checking that the second term is exactly $f'$ . $e^x(\sin x - \cos x)$ is not this pattern with $f = \sin x$ , since $f' = +\cos x$ . (It is the pattern with $f = -\cos x$ , giving $-e^x\cos x + C$ .)
Using it with $e^{2x}$ . The general version is $\int e^{kx}\left[kf(x) + f'(x)\right]dx = e^{kx}f(x) + C$ , so with $e^{2x}$ the bracket must be $2f + f'$ .

Where this comes up in exams

This pattern comes from integration by parts. JEE Main 2026 lists integration by substitution, by parts and by partial fractions, and the 2025-26 CBSE Class XII syllabus lists integration of a variety of functions by substitution, by partial fractions and by parts.

Checked against JEE (Main) 2026 syllabus (NTA) and CBSE Mathematics curriculum, Classes XI-XII (2025-26). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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