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Some integrals look endless by parts and impossible by substitution, yet finish in one line once you see that the bracket is a function plus its own derivative. This post shows why ∫eˣ[f(x) + f′(x)] dx = eˣf(x) + C works, how to spot the pattern when it is disguised, and how it extends to e^(ax) and e^(−x). Every answer is checked by differentiating.
Where is this in the syllabus?
NCERT Class 12 Integrals (reprint 2026-27) treats this as section 7.6.1, "Integral of the type ∫eˣ[f(x) + f′(x)]dx", inside integration by parts. The JEE (Main) 2026 syllabus lists "Integration by substitution, by parts and by partial fractions", the JEE (Advanced) 2026 brochure lists integration by parts, and the CBSE 2026-27 curriculum includes integration by parts for Class 12. Check the documents for your exam year.
Why does the rule work?
The quickest proof uses the product rule:
d/dx[eˣf(x)] = eˣf(x) + eˣf′(x) = eˣ[f(x) + f′(x)]
So eˣf(x) is an antiderivative. NCERT's route splits the integral into ∫eˣf(x) dx + ∫eˣf′(x) dx and integrates the first by parts, taking f(x) as the first function. That produces f(x)eˣ − ∫f′(x)eˣ dx, and the leftover integral cancels the second one exactly.
Worked examples
1. ∫eˣ(sin x + cos x) dx. Take f = sin x, so f′ = cos x. Answer: eˣ sin x + C. Check: d/dx(eˣ sin x) = eˣ sin x + eˣ cos x.
2. ∫eˣ(log x + 1/x) dx. Take f = log x, f′ = 1/x. Answer: eˣ log x + C.
3. ∫eˣ(x² + 2x) dx. Take f = x², f′ = 2x. Answer: x²eˣ + C. By parts this takes two rounds; the pattern takes one line.
4. ∫eˣ (x − 1)/x² dx. Split the fraction: (x − 1)/x² = 1/x − 1/x². With f = 1/x, f′ = −1/x². Answer: eˣ/x + C.
5. ∫eˣ (x + 1)/(x + 2)² dx. Write the numerator as (x + 2) − 1, so the integrand is eˣ[1/(x + 2) − 1/(x + 2)²]. With f = 1/(x + 2), f′ = −1/(x + 2)². Answer: eˣ/(x + 2) + C. Check: the derivative is eˣ/(x + 2) − eˣ/(x + 2)² = eˣ(x + 1)/(x + 2)².
6. A definite integral. ∫ from 1 to 2 of eˣ(1/x − 1/x²) dx = [eˣ/x] from 1 to 2 = e²/2 − e. With e² ≈ 7.3891 and e ≈ 2.7183, this is 3.6945 − 2.7183 ≈ 0.976.
What changes with e^(ax) or e^(−x)?
The product rule gives d/dx[e^(ax)f(x)] = e^(ax)[a f(x) + f′(x)]. So:
∫e^(ax)[a f(x) + f′(x)] dx = e^(ax)f(x) + C.
Example: ∫e^(2x)(2 tan x + sec²x) dx = e^(2x) tan x + C, since a = 2 and f = tan x.
For a = −1, the bracket becomes f′ − f. A useful form is ∫e^(−x)[f(x) − f′(x)] dx = −e^(−x)f(x) + C.
Example: ∫e^(−x)(cos x + sin x) dx. With f = cos x, f − f′ = cos x + sin x. Answer: −e^(−x) cos x + C. Check: d/dx(−e^(−x) cos x) = e^(−x) cos x + e^(−x) sin x.
How do you spot a disguised case?
Guess f from the "bigger" term and differentiate it. If the rest of the bracket appears, you are done.
Split fractions by writing the numerator as a multiple of the denominator plus a remainder, as in example 5.
Read the exponent. For e^(ax), you need a times f plus f′, not just f plus f′.
Pull out constants first. eˣ(3 sin x + 3 cos x) is 3eˣ(sin x + cos x).
Which mistakes cost marks?
Choosing the derivative as f. In example 1, taking f = cos x gives eˣ cos x, whose derivative is eˣ(cos x − sin x). The check catches it.
Ignoring the a in e^(ax). ∫e^(2x)(tan x + sec²x) dx does not fit the pattern.
Sign slips with e^(−x).
Reaching for parts twice when the pattern is there.
Forgetting + C in indefinite integrals.
A practice routine
Studying several versions of one idea helps. A review of worked-examples research by Atkinson, Derry, Renkl and Wortham (2000) concluded that effective instruction uses multiple examples for each problem type and varies the example formats within a type.
Pattern set (twice a week): six integrals, two plain, two with split fractions, two with e^(ax). Check each by differentiating.
Mixed set (weekly): add these to general integration by parts problems so you must decide when the shortcut applies.
Build your own: pick any f, compute f + f′, and give the integral to a friend.
Plan the rest: our integration practice plan spaces the other techniques across weeks.
If you can see the pattern but not why parts gives it, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Today, choose f(x) = tan⁻¹x, write out eˣ[f(x) + f′(x)], and integrate it in one line.
