3-1 Integration of the exponential and logarithmic functions: practice questions

Correct answer: (A) −2e^(−3x) + c

Explanation: Step 1: ∫ e^(kx) dx = (1/k)e^(kx) + c, here k = −3. Step 2: ∫ 6e^(−3x) dx = 6 × (1/(−3)) e^(−3x) + c. Step 3: ∴ −2e^(−3x) + c

Correct answer: (B) eˣ + c

Explanation: Step 1: Recall d/dx(eˣ) = eˣ. Step 2: Hence ∫eˣ dx = eˣ + c. ∴ ∫eˣ dx = eˣ + c

Correct answer: (D) eˣ + ln|x| + c

Explanation: Step 1: Integrate each term: ∫ eˣ dx = eˣ. Step 2: ∫ (1/x) dx = ln|x|. Step 3: ∴ eˣ + ln|x| + c

Correct answer: (D) eˣ + 4e^(−x) + c

Explanation: Step 1: Divide each term by eˣ: (e^(2x) − 4)/eˣ = eˣ − 4e^(−x). Step 2: ∫ eˣ dx = eˣ and ∫ −4e^(−x) dx = −4 × (1/(−1)) e^(−x) = 4e^(−x). Step 3: ∴ eˣ + 4e^(−x) + c

Correct answer: (A) (1/3)e^(3x) + c

Explanation: Step 1: Use ∫e^(ax) dx = (1/a)e^(ax) + c, with a = 3. Step 2: = (1/3)e^(3x) + c. ∴ ∫e^(3x) dx = (1/3)e^(3x) + c

Correct answer: (C) (1/2)ln|2x+1| + c

Explanation: Step 1: Use ∫dx/(ax+b) = (1/a)ln|ax+b| + c, with a = 2. Step 2: = (1/2)ln|2x+1| + c. ∴ ∫1/(2x+1) dx = (1/2)ln|2x+1| + c

Correct answer: (C) (1/2)e^(2x+1) + c

Explanation: Step 1: Use ∫e^(ax+b) dx = (1/a)e^(ax+b) + c, with a = 2. Step 2: = (1/2)e^(2x+1) + c. ∴ ∫e^(2x+1) dx = (1/2)e^(2x+1) + c

Correct answer: (C) −ln|cos x| + c

Explanation: Step 1: Write tan x = sin x/cos x; let u = cos x ⇒ du = −sin x dx. Step 2: ∫(sin x/cos x) dx = −∫du/u = −ln|u|. Step 3: = −ln|cos x| + c. ∴ ∫tan x dx = −ln|cos x| + c

Correct answer: (A) (3/2) ln|ln x| + c

Explanation: Step 1: ln x² = 2 ln x, so 3/(x ln x²) = (3/2) × (1/x)/ln x. Step 2: Let f(x) = ln x, then f′(x) = 1/x, so the integrand is (3/2) f′(x)/f(x). Step 3: ∫ f′(x)/f(x) dx = ln|f(x)| + c. Step 4: ∴ (3/2) ln|ln x| + c

Correct answer: (C) e^(2x)/2 − eˣ + x + c

Explanation: Step 1: Sum of two cubes: e^(3x) + 1 = (eˣ + 1)(e^(2x) − eˣ + 1). Step 2: So the integrand = e^(2x) − eˣ + 1. Step 3: ∫ e^(2x) dx = e^(2x)/2, ∫ eˣ dx = eˣ, ∫ 1 dx = x. Step 4: ∴ e^(2x)/2 − eˣ + x + c