Integration by substitution and by parts practice questions.

Using integration by parts, determine ∫ x cos x dx.

• x sin x - cos x + C
• x sin x + cos x + C
• -x sin x + cos x + C
• x sin x - sin x + C

Answer: x sin x + cos x + C

With u = x, dv = cos x dx: ∫ x cos x dx = x sin x - ∫ sin x dx = x sin x + cos x + C. Differentiating confirms it: sin x + x cos x - sin x = x cos x.

Using the substitution u = x² + 1, determine ∫ x√(x² + 1) dx.

• (1/3)(x² + 1)^(3/2) + C
• (1/2)(x² + 1)^(3/2) + C
• (2/3)(x² + 1)^(3/2) + C
• (1/3)(x² + 1)^(1/2) + C

Answer: (1/3)(x² + 1)^(3/2) + C

du = 2x dx, so the integral becomes (1/2)∫ √u du = (1/2)·(2/3)u^(3/2) = (1/3)(x² + 1)^(3/2) + C.

Determine ∫ ln x dx.

• (ln x)²/2 + C
• x ln x + x + C
• x ln x - x + C
• 1/x + C

Answer: x ln x - x + C

Take u = ln x, dv = dx: ∫ ln x dx = x ln x - ∫ x·(1/x) dx = x ln x - x + C.

Using integration by parts twice, determine ∫ x² eˣ dx.

• (x² + 2x + 2)eˣ + C
• (x² - 2x - 2)eˣ + C
• (x²/3)eˣ + C
• (x² - 2x + 2)eˣ + C

Answer: (x² - 2x + 2)eˣ + C

Two applications of parts give x²eˣ - 2xeˣ + 2eˣ + C = (x² - 2x + 2)eˣ + C; differentiating returns (2x - 2)eˣ + (x² - 2x + 2)eˣ = x²eˣ.

Using the substitution u = cos x, determine ∫ tan x dx.

• ln|cos x| + C
• -ln|cos x| + C
• ln|sin x| + C
• -ln|sin x| + C

Answer: -ln|cos x| + C

tan x = sin x/cos x and du = -sin x dx, so the integral is -∫ du/u = -ln|u| + C = -ln|cos x| + C (equivalently ln|sec x| + C).

The substitution u = x² + 1 is used to evaluate ∫ from 0 to 1 of 2x(x² + 1)³ dx. State the new limits of integration in terms of u.

• u = 0 to u = 1
• u = 0 to u = 2
• u = 1 to u = 2
• u = 1 to u = 4

Answer: u = 1 to u = 2

The limits transform via u = x² + 1: when x = 0, u = 1; when x = 1, u = 2. Forgetting to convert limits is a classic error.