u so that its derivative (up to a constant) appears in the integrand, then rewrite the whole integral in u.g'/g → ln|g|, eᵍ g' → eᵍ, aˣ → aˣ/ln a, and the inverse-trig forms arctan and arcsin — and decide when Section 2's calculator can check an antiderivative.When an integrand looks like a function of something times the derivative of that something, substitution collapses it into one variable:
∫ f(g(x))·g'(x) dx → let u = g(x), du = g'(x) dx → ∫ f(u) du
The steps: (1) choose u as the inside function whose derivative also appears; (2) compute du = u'(x) dx and solve for the grouping you have; (3) rewrite the integral entirely in u — if a stray x survives, your u was wrong; (4) integrate in u; (5) back-substitute (indefinite, + C) or change limits (definite).
A missing constant factor can always be supplied, because constants pass freely through an integral. For ∫ x(x² + 1)³ dx, let u = x² + 1, so du = 2x dx, giving x dx = (1/2) du:
∫ x(x² + 1)³ dx = (1/2)∫ u³ du = u⁴/8 + C = (x² + 1)⁴/8 + C
A missing variable factor cannot be supplied — that means u was chosen incorrectly. So ∫ (x² + 1)³ dx (no x factor) is not a direct substitution.
∫ (g)ⁿ g' dx = (g)ⁿ⁺¹/(n+1) + C (n ≠ −1).∫ eᵍ g' dx = eᵍ + C.∫ (g'/g) dx = ln|g| + C — the numerator is the derivative of the denominator. Keep the absolute value unless g > 0 is guaranteed.∫ aˣ dx = aˣ/ln a + C (a > 0, a ≠ 1)
∫ dx/(a² + x²) = (1/a) arctan(x/a) + C
∫ dx/√(a² − x²) = arcsin(x/a) + C
The tells: a plain sum a² + x² in a denominator → arctan (with a 1/a factor). A square root over a difference √(a² − x²) → arcsin (no 1/a factor). Two clean examples: ∫ dx/(x² + 9) = (1/3) arctan(x/3) + C and ∫ dx/√(9 − x²) = arcsin(x/3) + C.
For ∫ₐᵇ f(g(x)) g'(x) dx, the cleaner method is to convert the x-limits into u-limits: x = a → u = g(a), x = b → u = g(b), then never return to x.
∫₀^{π/2} sin x cos x dx (u = sin x, du = cos x dx; limits 0→0, π/2→1)
= ∫₀¹ u du = [u²/2]₀¹ = 1/2
The trap is doing both — changing the limits and also plugging the original x-limits into a u-expression. Pick one method and stay consistent.
Section 1 has no calculator, so these antiderivatives must be produced by hand. In Section 2, the built-in graphing calculator can evaluate a definite integral numerically — a fast way to confirm an antiderivative you found by hand, or to land on a best-approximation decimal answer (e.g. ∫₀² x eˣ² dx ≈ 26.799) when the exact form is not among the choices.
u so du (its derivative) appears up to a constant factor; a leftover x means the wrong u.g'/g → ln|g|, eᵍ g' → eᵍ, aˣ → aˣ/ln a.1/a); root over a difference → arcsin (no 1/a).u-values — do not also revert to x.∫ (2x + 1)⁵ dx =1. B — Correct: u = 2x + 1, du = 2 dx, so dx = (1/2) du, giving (2x + 1)⁶/12 + C. A forgot the 1/2 from du. C forgot to divide by the new exponent 6. D differentiated. E kept only the 1/2 and dropped the 1/6. Fix: carry both the 1/(n+1) and the constant from du.
∫ x²(x³ − 1)⁴ dx =2. D — Correct: u = x³ − 1, du = 3x² dx, so x² dx = (1/3) du, giving (x³ − 1)⁵/15 + C. A forgot the 1/3. B and C mishandle the 1/3 constant. E divided by the wrong number. Fix: solve du for exactly the grouping present (x² dx = (1/3) du).
∫ e^{5x} dx =3. D — Correct: u = 5x, du = 5 dx, so (1/5)e^{5x} + C. A forgot to divide by 5. B multiplied by 5 (that is differentiation). C and E illegally divided by the variable. Fix: ∫e^{kx} dx = (1/k)e^{kx} + C.
∫ x/(x² + 4) dx =4. C — Correct: log form, u = x² + 4, du = 2x dx, so x dx = (1/2) du, giving (1/2) ln(x² + 4) + C. A forgot the 1/2. B doubled instead of halving. D treated it as a power rather than a log. E used the arctan form, which needs a constant numerator, not x. Fix: numerator = (half of) the derivative of the denominator ⇒ logarithm with the 1/2.
∫ dx/(25 + x²) =5. B — Correct: ∫ dx/(a² + x²) = (1/a) arctan(x/a) with a = 5. A dropped the 1/5. C inverted the structure. D used arcsin instead of arctan (plain sum, no root → arctan). E squared the 1/a factor. Fix: plain sum in the denominator → (1/a) arctan(x/a).
∫ dx/√(9 − x²) =6. A — Correct: ∫ dx/√(a² − x²) = arcsin(x/a) with a = 3; this form has no 1/a factor out front. B added a spurious 1/3. C used arctan (root over a difference → arcsin). D mistook it for a log. E combined both errors. Fix: root over a difference → arcsin(x/a), no leading fraction.
∫₀^{π/2} sin x cos x dx =7. E — Correct: u = sin x, du = cos x dx, limits 0→0 and π/2→1, so ∫₀¹ u du = 1/2. A ignored the substitution. B and C are unrelated values. D forgot the 1/2 from ∫u du = u²/2. Fix: change the limits to u-values and evaluate u²/2.
∫₀² x e^{x²} dx is best approximated by8. C — Correct: u = x², x dx = (1/2) du, limits 0→0 and 2→4, so (1/2)(e⁴ − e⁰) = (e⁴ − 1)/2 ≈ 26.799. A is e⁴ − 1 (forgot the 1/2). E is e⁴ (forgot both the −1 and the 1/2). B is a near-miss decimal. D confuses this with a sin·cos result. Fix: keep the 1/2 from du, then evaluate the exponentials.
∫₁^{e} (ln x)/x dx =9. D — Correct: u = ln x, du = (1/x) dx, limits 1→0 and e→1, so ∫₀¹ u du = 1/2. A forgot the 1/2. B and E ignore the substitution. C mis-evaluates the limits. Fix: (ln x)·(1/x) is u du with u = ln x.
∫ (ln x)/x dx =10. E — Correct: u = ln x, du = (1/x) dx, so ∫ u du = (ln x)²/2 + C. A and B are derivative-style guesses. C misapplies the g'/g log pattern (here the integrand is u du, not g'/g). D forgot the 1/2. Fix: recognize (1/x) dx as du for u = ln x.
∫ x/√(x² + 16) dx =11. A — Correct: u = x² + 16, x dx = (1/2) du, so (1/2)∫ u^{−1/2} du = √(x² + 16) + C. B and C mishandle the constant. D treated it as a log. E used arcsin, the wrong form entirely. Fix: ∫ u^{−1/2} du = 2√u, and the (1/2)(2) = 1 cancels cleanly.
u = x² + 1 immediately effective, so that du appears up to a constant factor?12. E — Correct: ∫ x(x² + 1)⁴ dx has the factor x, and du = 2x dx gives x dx = (1/2) du, so the integral becomes fully in u. A has no x factor — a stray x remains after substituting. B is an arctan form, not a power substitution. C has x², not x, leaving a leftover x. D has x in the denominator, not as g'. Fix: substitution needs the derivative of u present up to a constant — here that derivative is 2x, so you need exactly one factor of x.
1. B — Correct: u = 2x + 1, du = 2 dx, so dx = (1/2) du, giving (2x + 1)⁶/12 + C. A forgot the 1/2 from du. C forgot to divide by the new exponent 6. D differentiated. E kept only the 1/2 and dropped the 1/6. Fix: carry both the 1/(n+1) and the constant from du.
2. D — Correct: u = x³ − 1, du = 3x² dx, so x² dx = (1/3) du, giving (x³ − 1)⁵/15 + C. A forgot the 1/3. B and C mishandle the 1/3 constant. E divided by the wrong number. Fix: solve du for exactly the grouping present (x² dx = (1/3) du).
3. D — Correct: u = 5x, du = 5 dx, so (1/5)e^{5x} + C. A forgot to divide by 5. B multiplied by 5 (that is differentiation). C and E illegally divided by the variable. Fix: ∫e^{kx} dx = (1/k)e^{kx} + C.
4. C — Correct: log form, u = x² + 4, du = 2x dx, so x dx = (1/2) du, giving (1/2) ln(x² + 4) + C. A forgot the 1/2. B doubled instead of halving. D treated it as a power rather than a log. E used the arctan form, which needs a constant numerator, not x. Fix: numerator = (half of) the derivative of the denominator ⇒ logarithm with the 1/2.
5. B — Correct: ∫ dx/(a² + x²) = (1/a) arctan(x/a) with a = 5. A dropped the 1/5. C inverted the structure. D used arcsin instead of arctan (plain sum, no root → arctan). E squared the 1/a factor. Fix: plain sum in the denominator → (1/a) arctan(x/a).
6. A — Correct: ∫ dx/√(a² − x²) = arcsin(x/a) with a = 3; this form has no 1/a factor out front. B added a spurious 1/3. C used arctan (root over a difference → arcsin). D mistook it for a log. E combined both errors. Fix: root over a difference → arcsin(x/a), no leading fraction.
7. E — Correct: u = sin x, du = cos x dx, limits 0→0 and π/2→1, so ∫₀¹ u du = 1/2. A ignored the substitution. B and C are unrelated values. D forgot the 1/2 from ∫u du = u²/2. Fix: change the limits to u-values and evaluate u²/2.
8. C — Correct: u = x², x dx = (1/2) du, limits 0→0 and 2→4, so (1/2)(e⁴ − e⁰) = (e⁴ − 1)/2 ≈ 26.799. A is e⁴ − 1 (forgot the 1/2). E is e⁴ (forgot both the −1 and the 1/2). B is a near-miss decimal. D confuses this with a sin·cos result. Fix: keep the 1/2 from du, then evaluate the exponentials.
9. D — Correct: u = ln x, du = (1/x) dx, limits 1→0 and e→1, so ∫₀¹ u du = 1/2. A forgot the 1/2. B and E ignore the substitution. C mis-evaluates the limits. Fix: (ln x)·(1/x) is u du with u = ln x.
10. E — Correct: u = ln x, du = (1/x) dx, so ∫ u du = (ln x)²/2 + C. A and B are derivative-style guesses. C misapplies the g'/g log pattern (here the integrand is u du, not g'/g). D forgot the 1/2. Fix: recognize (1/x) dx as du for u = ln x.
11. A — Correct: u = x² + 16, x dx = (1/2) du, so (1/2)∫ u^{−1/2} du = √(x² + 16) + C. B and C mishandle the constant. D treated it as a log. E used arcsin, the wrong form entirely. Fix: ∫ u^{−1/2} du = 2√u, and the (1/2)(2) = 1 cancels cleanly.
12. E — Correct: ∫ x(x² + 1)⁴ dx has the factor x, and du = 2x dx gives x dx = (1/2) du, so the integral becomes fully in u. A has no x factor — a stray x remains after substituting. B is an arctan form, not a power substitution. C has x², not x, leaving a leftover x. D has x in the denominator, not as g'. Fix: substitution needs the derivative of u present up to a constant — here that derivative is 2x, so you need exactly one factor of x.