+ C.n = −1 logarithm case), plus the exponential and trigonometric antiderivative rules, rewriting integrands first when needed.+ CA function F is an antiderivative of f if F'(x) = f(x). Because the derivative of any constant is 0, if F is one antiderivative then F(x) + C is an antiderivative for every constant C, and these are the only ones. So antidifferentiation never returns a single function — it returns a whole one-parameter family. The indefinite integral names that family:
∫ f(x) dx = F(x) + C, where F'(x) = f(x)
The dx marks the variable of integration; the + C is the constant of integration and belongs on every indefinite integral. On the CLEP exam, dropping it turns a right answer into a wrong one.
Each rule below is a differentiation rule read backward. Verify any of them by differentiating the right-hand side.
| Integral | Antiderivative |
|---|---|
∫ k dx |
kx + C |
∫ xⁿ dx (n ≠ −1) |
xⁿ⁺¹/(n+1) + C |
∫ (1/x) dx |
ln|x| + C |
∫ eˣ dx |
eˣ + C |
∫ aˣ dx (a > 0, a ≠ 1) |
aˣ/ln a + C |
∫ cos x dx |
sin x + C |
∫ sin x dx |
−cos x + C |
∫ sec²x dx |
tan x + C |
Power rule: add one to the exponent, then divide by the new exponent. It works for negative and fractional exponents too, once you rewrite radicals and quotients as powers (√x = x^(1/2), 1/x³ = x⁻³).
The n = −1 special case. The power rule fails for ∫ x⁻¹ dx because adding one gives exponent 0 and you would divide by 0. That single case is patched by the logarithm: ∫ (1/x) dx = ln|x| + C. The absolute value matters — 1/x is defined for negative x, and d/dx[ln|x|] = 1/x on both sides of 0.
Antidifferentiation is linear: constants pull out front, and sums integrate term by term. There is no product rule and no quotient rule for integrals — if you face a product or a quotient, you must rewrite it as a sum of power/trig/exponential terms first. Common moves: split a fraction over its denominator ((x³ − 4x + 1)/x² = x − 4·x⁻¹ + x⁻²), expand a product ((x² + 1)² = x⁴ + 2x² + 1), and write radicals as powers.
Worked: ∫ (√x − 5/x + eˣ) dx = (2/3)x^(3/2) − 5 ln|x| + eˣ + C. The middle term is the n = −1 trap — 5/x integrates to a logarithm, not a power.
The most reliable check in integration: differentiate your answer. If you recover the integrand, you are right. Do this whenever you have the seconds.
A particular antiderivative is one member of the family, fixed by an initial condition F(a) = b: integrate to get F(x) + C, substitute the condition, solve for C. Antiderivatives also recover motion. Given acceleration a(t), integrate once (using v(0)) to get velocity, and again (using s(0)) to get position:
v(t) = ∫ a(t) dt s(t) = ∫ v(t) dt
One initial condition is needed per integration. Because the integral of a rate gives accumulated change, if s(t) is position then s(b) = s(a) + ∫ₐᵇ v(t) dt.
∫ f(x) dx = F(x) + C; every indefinite integral carries + C.∫ (1/x) dx = ln|x| + C.∫ sin x dx = −cos x + C (mind the sign); ∫ aˣ dx = aˣ/ln a + C (mind the 1/ln a).C; acceleration → velocity → position, one condition each.∫ (√x − 5/x + eˣ) dx =1. C — Correct: √x = x^(1/2) integrates to (2/3)x^(3/2), −5/x to −5 ln|x|, eˣ to eˣ. A differentiated the √x term instead of integrating it. B inverted the exponent fraction (3/2 instead of 2/3). D dropped the + C. E applied the power rule to −5/x instead of using the n = −1 log rule. Fix: rewrite radicals as powers, and treat any lone 1/x term as a logarithm.
∫ (6x² − 4/x³ + 2) dx =2. B — Correct: 6x² → 2x³, −4x⁻³ → −4·x⁻²/(−2) = +2/x², 2 → 2x. A dropped the sign flip when dividing by the new exponent −2. C differentiated the middle term. D failed to integrate the constant 2 (left it as 2 instead of 2x). E forgot to divide the first term by its new exponent. Fix: divide by the new exponent every term, and watch the sign a negative exponent introduces.
∫ (4 cos x − 3 sec²x + 2eˣ) dx =3. A — Correct: ∫4 cos x = 4 sin x, ∫−3 sec²x = −3 tan x, ∫2eˣ = 2eˣ. B put the wrong sign on the cosine term (∫cos = +sin). C used sec x instead of tan x (the antiderivative of sec²x is tan x). D dropped + C. E left 4 cos x unintegrated. Fix: ∫cos = sin, ∫sec²x = tan x; verify by differentiating.
f'(x) = 3x² + 2 and f(0) = 4, then f(x) =4. B — Correct: f(x) = x³ + 2x + C, and f(0) = C = 4. A found the family but ignored the initial condition. C left C unsolved despite the given value. D differentiated. E mis-solved for C (used 2 instead of 4). Fix: an initial-condition problem demands a number for C — substitute and solve.
a(t) = 6t − 4 with v(0) = 2. Its velocity function is5. A — Correct: v(t) = ∫(6t − 4) dt = 3t² − 4t + C, and v(0) = C = 2. B ignored the initial condition. C left C unsolved. D differentiated the acceleration. E forgot to divide 6t by the new exponent 2. Fix: integrate acceleration, then use v(0) to fix the constant.
f(x) = 1/x?6. D — Correct: any function ln|x| + C is an antiderivative of 1/x; here C = −5. A and B are derivative-style guesses (d/dx[1/x²] ≠ 1/x). C is undefined — the power rule fails at n = −1. E has derivative eˣ, not 1/x. Fix: the antiderivative of 1/x is ln|x| plus any constant.
∫ (x³ − 4x + 1)/x² dx =7. D — Correct: split first, (x³ − 4x + 1)/x² = x − 4·(1/x) + x⁻², giving x²/2 − 4 ln|x| − 1/x + C. A mis-integrated the x term. B put the wrong sign on the log term. C put the wrong sign on the −1/x term (∫x⁻² = −1/x). E integrated the 1/x term as a power instead of a log. Fix: split the fraction, then treat each piece by its own rule — the 1/x piece is a logarithm.
∫ (x² + 1)² dx =8. E — Correct: expand (x² + 1)² = x⁴ + 2x² + 1, then integrate to x⁵/5 + (2/3)x³ + x + C. A invented a "chain rule for integrals." B failed to divide by the new exponents. C differentiated. D dropped + C. Fix: there is no chain rule for integrals — expand the product first.
F and G are both antiderivatives of the same continuous function f on an interval. Which statement must be true?9. E — Correct: two antiderivatives of the same function differ by a constant, so F − G is constant. A and C are too strong — the constant need not be 0. B and D contradict the fact that both have derivative f. Fix: same derivative on an interval ⇒ the functions differ by a constant.
v(t) = 3t² − 6t and position s(0) = 4. Then s(2) =10. E — Correct: s(t) = ∫(3t² − 6t) dt = t³ − 3t² + C; s(0) = C = 4, so s(2) = 8 − 12 + 4 = 0. A is the displacement s(2) − s(0), not the position. B keeps only the constant. C is t³ alone at t = 2. D is an arithmetic slip. Fix: find s(t) with its constant, then evaluate at the requested time.
∫ (2ˣ + 4 cos x) dx =11. C — Correct: ∫2ˣ = 2ˣ/ln 2 and ∫4 cos x = 4 sin x. A forgot the 1/ln 2. B multiplied by ln 2 (that is differentiation). D applied the power rule to a variable exponent. E put the wrong sign on the sine term. Fix: ∫aˣ dx = aˣ/ln a + C.
∫ (1/x) dx =12. A — Correct: the n = −1 special case, ∫(1/x) dx = ln|x| + C. B drops the absolute value, which is needed for negative x. C differentiated. D shows the power rule failing (division by zero). E is unrelated. Fix: memorize ∫(1/x) dx = ln|x| + C, absolute value included.
1. C — Correct: √x = x^(1/2) integrates to (2/3)x^(3/2), −5/x to −5 ln|x|, eˣ to eˣ. A differentiated the √x term instead of integrating it. B inverted the exponent fraction (3/2 instead of 2/3). D dropped the + C. E applied the power rule to −5/x instead of using the n = −1 log rule. Fix: rewrite radicals as powers, and treat any lone 1/x term as a logarithm.
2. B — Correct: 6x² → 2x³, −4x⁻³ → −4·x⁻²/(−2) = +2/x², 2 → 2x. A dropped the sign flip when dividing by the new exponent −2. C differentiated the middle term. D failed to integrate the constant 2 (left it as 2 instead of 2x). E forgot to divide the first term by its new exponent. Fix: divide by the new exponent every term, and watch the sign a negative exponent introduces.
3. A — Correct: ∫4 cos x = 4 sin x, ∫−3 sec²x = −3 tan x, ∫2eˣ = 2eˣ. B put the wrong sign on the cosine term (∫cos = +sin). C used sec x instead of tan x (the antiderivative of sec²x is tan x). D dropped + C. E left 4 cos x unintegrated. Fix: ∫cos = sin, ∫sec²x = tan x; verify by differentiating.
4. B — Correct: f(x) = x³ + 2x + C, and f(0) = C = 4. A found the family but ignored the initial condition. C left C unsolved despite the given value. D differentiated. E mis-solved for C (used 2 instead of 4). Fix: an initial-condition problem demands a number for C — substitute and solve.
5. A — Correct: v(t) = ∫(6t − 4) dt = 3t² − 4t + C, and v(0) = C = 2. B ignored the initial condition. C left C unsolved. D differentiated the acceleration. E forgot to divide 6t by the new exponent 2. Fix: integrate acceleration, then use v(0) to fix the constant.
6. D — Correct: any function ln|x| + C is an antiderivative of 1/x; here C = −5. A and B are derivative-style guesses (d/dx[1/x²] ≠ 1/x). C is undefined — the power rule fails at n = −1. E has derivative eˣ, not 1/x. Fix: the antiderivative of 1/x is ln|x| plus any constant.
7. D — Correct: split first, (x³ − 4x + 1)/x² = x − 4·(1/x) + x⁻², giving x²/2 − 4 ln|x| − 1/x + C. A mis-integrated the x term. B put the wrong sign on the log term. C put the wrong sign on the −1/x term (∫x⁻² = −1/x). E integrated the 1/x term as a power instead of a log. Fix: split the fraction, then treat each piece by its own rule — the 1/x piece is a logarithm.
8. E — Correct: expand (x² + 1)² = x⁴ + 2x² + 1, then integrate to x⁵/5 + (2/3)x³ + x + C. A invented a "chain rule for integrals." B failed to divide by the new exponents. C differentiated. D dropped + C. Fix: there is no chain rule for integrals — expand the product first.
9. E — Correct: two antiderivatives of the same function differ by a constant, so F − G is constant. A and C are too strong — the constant need not be 0. B and D contradict the fact that both have derivative f. Fix: same derivative on an interval ⇒ the functions differ by a constant.
10. E — Correct: s(t) = ∫(3t² − 6t) dt = t³ − 3t² + C; s(0) = C = 4, so s(2) = 8 − 12 + 4 = 0. A is the displacement s(2) − s(0), not the position. B keeps only the constant. C is t³ alone at t = 2. D is an arithmetic slip. Fix: find s(t) with its constant, then evaluate at the requested time.
11. C — Correct: ∫2ˣ = 2ˣ/ln 2 and ∫4 cos x = 4 sin x. A forgot the 1/ln 2. B multiplied by ln 2 (that is differentiation). D applied the power rule to a variable exponent. E put the wrong sign on the sine term. Fix: ∫aˣ dx = aˣ/ln a + C.
12. A — Correct: the n = −1 special case, ∫(1/x) dx = ln|x| + C. B drops the absolute value, which is needed for negative x. C differentiated. D shows the power rule failing (division by zero). E is unrelated. Fix: memorize ∫(1/x) dx = ln|x| + C, absolute value included.