11 Practice questions with full solutions

Cover the solution, attempt it properly on paper, then reveal. They increase in difficulty.

Set A, indefinite integrals

① ∫ 3x²(x³ + 4)⁷ dx Reveal Let u = x³+4, du = 3x² dx. → ∫u⁷du = u⁸/8 + c.
= (x³ + 4)⁸ / 8 + c ② ∫ e^(5x) dx Reveal Let u = 5x, du = 5dx ⟹ dx = ⅕du. → ⅕∫e^u du = ⅕e^u + c.
= ⅕ e^(5x) + c ③ ∫ x/(x² + 7) dx Reveal Let u = x²+7, du = 2x dx ⟹ x dx = ½du. → ½∫(1/u)du = ½ln|u| + c.
= ½ ln(x² + 7) + c (no modulus needed since x²+7 > 0) ④ ∫ cos x · e^(sin x) dx Reveal Let u = sin x, du = cos x dx. → ∫e^u du = e^u + c.
= e^(sin x) + c ⑤ ∫ (ln x)³ / x dx Reveal Let u = ln x, du = (1/x)dx. → ∫u³du = u⁴/4 + c.
= (ln x)⁴ / 4 + c ⑥ ∫ x²√(x³ − 1) dx Reveal Let u = x³−1, du = 3x²dx ⟹ x²dx = ⅓du. → ⅓∫u^(1/2)du = ⅓·(2/3)u^(3/2) + c.
= (2/9)(x³ − 1)^(3/2) + c

Set B, definite integrals (change your limits!)

⑦ ∫₀¹ 2x(x² + 3)⁴ dx Reveal u = x²+3, du = 2x dx. Limits: x=0→u=3; x=1→u=4.
∫₃⁴u⁴du = [u⁵/5]₃⁴ = (1024 − 243)/5 = 781/5.
= 156.2 ⑧ ∫₁^e (ln x)/x dx Reveal u = ln x, du = (1/x)dx. Limits: x=1→u=ln1=0; x=e→u=ln e=1.
∫₀¹u du = [u²/2]₀¹ = ½ − 0.
= ½ ⑨ ∫₀² x·e^(x²) dx Reveal u = x², x dx = ½du. Limits: x=0→u=0; x=2→u=4.
½∫₀⁴e^u du = ½[e^u]₀⁴ = ½(e⁴ − 1).
= ½(e⁴ − 1) ≈ 26.8 ⑩ ∫₀^(π/2) sin²x · cos x dx Reveal u = sin x, du = cos x dx. Limits: x=0→u=0; x=π/2→u=sin(π/2)=1.
∫₀¹u²du = [u³/3]₀¹ = ⅓.
= ⅓

Set C, harder / rearrangement needed

⑪ ∫ x(x + 5)⁴ dx Reveal u = x+5 ⟹ x = u−5, du = dx.
∫(u−5)u⁴du = ∫(u⁵ − 5u⁴)du = u⁶/6 − u⁵ + c.
= (x+5)⁶/6 − (x+5)⁵ + c ⑫ ∫ (2x + 1)/(x² + x + 4) dx Reveal u = x²+x+4, du = (2x+1)dx: exactly the numerator.
∫(1/u)du = ln|u| + c.
= ln|x² + x + 4| + c ⑬ ∫₀¹ x/√(x + 3) dx Reveal u = x+3 ⟹ x = u−3, du = dx. Limits: x=0→u=3; x=1→u=4.
(2/3)u^(3/2) − 6u^(1/2), so (16/3 − 12) + 4√3 ≈ 0.261

Practice for this standard covers the whole topic — flashcards and quiz.