Subiectul al III-lea, 2., toate variantele

2.a) ∫₀¹ (f(x)/eˣ) dx = ∫₀¹ (x² + x) dx = [x³/3]₀¹ + [x²/2]₀¹ = - 3p = 1/3 + 1/2 = 5/6 - 2p

2.b) ∫₁² (f(x)/x) dx = ∫₁² (x + 1)(eˣ)′ dx = [xeˣ]₁² = - 3p = 2e² − e = e(2e − 1) - 2p

2.c) ∫₂⁴ (xf(x)/f(x − 1)) dx = e∫₂⁴ ((x² + x)/(x − 1)) dx = e∫₂⁴ (x + 2 + 2/(x − 1)) dx = e[x²/2 + 2x + 2ln(x − 1)]₂⁴ = 2e(5 + ln 3) - 3p ae(5 + ln 3) = 2e(5 + ln 3), de unde obținem a = 2 - 2p

∫₀³ f(x)/√(x + 1) dx = ∫₀³ (2x² + 4x) dx = [2x³/3]₀³ + [2x²]₀³ = - 3p = 18 + 18 = 36 - 2p

∫₃⁸ (x² + 2x)/f(x) dx = ∫₃⁸ 1/(2√(x + 1)) dx = ∫₃⁸ (x + 1)′/(2√(x + 1)) dx = [√(x + 1)]₃⁸ = - 3p = 3 − 2 = 1 - 2p

Reguli generale:

2. c) g(x) = 2/(x² + 2x), x ∈ (0, +∞), deci A = ∫₁⁴ |g(x)| dx = ∫₁⁴ 2/[x(x + 2)] dx = ∫₁⁴ (1/x − 1/(x + 2)) dx = - 3p = [ln x]₁⁴ − [ln(x + 2)]₁⁴ = ln 2 - 2p

∫₂³ (f(x) − (2 ln x)/x) dx = ∫₂³ 6x dx = [3x²]₂³ = - 3p = 27 − 12 = 15 - 2p

∫₁ᵉ (f(x) − 6x) dx = ∫₁ᵉ (2 ln x)/x dx = ∫₁ᵉ 2 ln x (ln x)′ dx = [ln² x]₁ᵉ = - 3p = ln² e − ln² 1 = 1 - 2p

∫₁ᵉ² (f(x)/x + f′(x) ln x) dx = ∫₁ᵉ² (f(x) ln x)′ dx = [f(x) ln x]₁ᵉ² = f(e²) ln e² = 2f(e²) - 3p 2f(e²) = a f(e²) și, cum f(e²) ≠ 0, obținem a = 2 - 2p

2.a) ∫₀² (f(x) − 3x − 3) dx = ∫₀² x³ dx = [x⁴/4]₀² = - 3p = 4 − 0 = 4 - 2p

b) ∫₀¹ 1/(f(x) − x³)² dx = (1/3) ∫₀¹ (3x + 3)′/(3x + 3)² dx = −(1/3) · [1/(3x + 3)]₀¹ = - 3p = −1/18 + 1/9 = 1/18 - 2p

c) ∫₁ᵉ ((f(x) − 3)/x²) · ln x dx = ∫₁ᵉ (x²/2)′ · ln x dx + ∫₁ᵉ (3 ln x)/x dx = [(x²/2) · ln x − x²/4]₁ᵉ + [3 ln² x/2]₁ᵉ = (e² + 7)/4 - 3p (e² + m)/4 = (e² + 7)/4, de unde obținem m = 7 - 2p

2.a) ∫₂⁴ f(x)√x dx = ∫₂⁴ (x−1) dx = [x²/2]₂⁴ − [x]₂⁴ = - 3p = 6−2 = 4 - 2p

b) ∫₁⁴ f(x) dx = ∫₁⁴ (√x − 1/√x) dx = [2x√x/3]₁⁴ − [2√x]₁⁴ = - 3p = 14/3−2 = 8/3 - 2p

c) g(x) = √(2x)/(x−1), x ∈ [2,3], deci V = π∫₂³ (g(x))² dx = 2π∫₂³ x/(x−1)² dx = - 2p = 2π∫₂³ (1/(x−1) + 1/(x−1)²) dx = 2π([ln(x−1)]₂³ − [1/(x−1)]₂³) = π ln(4e) - 3p

2. a) ∫₂⁴ (f(x) − 1/√(x + 3)) dx = ∫₂⁴ 3x dx = [3x²/2]₂⁴ = - 3p = 48/2 − 12/2 = 18 - 2p

2. b) ∫₁⁶ (f(x) − 3x) dx = ∫₁⁶ 1/√(x + 3) dx = 2∫₁⁶ (x + 3)′/(2√(x + 3)) dx = [2√(x + 3)]₁⁶ = - 3p = 6 − 4 = 2 - 2p

2. c) ∫₋₂¹ [1/(x + 3)](f(x) − 1/√(x + 3)) dx = ∫₋₂¹ (3 − 9/(x + 3)) dx = [3x]₋₂¹ − [9 ln(x + 3)]₋₂¹ = 9(1 − ln 4) - 3p 9(1 − 2 ln 2) = 9(a − 2 ln 2), de unde obținem a = 1 - 2p

∫₀² (x + 1)f(x) dx = ∫₀² (x + 3) dx = [x²/2 + 3x]₀² = - 3p = 2 + 6 = 8 - 2p

∫₀¹ f(x) dx = ∫₀¹ (x + 3)/(x + 1) dx = ∫₀¹ (1 + 2/(x + 1)) dx = [x + 2ln(x + 1)]₀¹ = - 3p = 1 + 2ln 2 − 2ln 1 = 1 + 2ln 2 - 2p

∫₁² (x² − 1)eˣf(x) dx = ∫₁² (x² + 2x − 3)eˣ dx = [(x² − 3)eˣ]₁² = e² + 2e - 3p e(e + a) = e² + 2e, de unde obținem a = 2 - 2p

∫₀³(f(x)−3x) dx=∫₀³(x²+1) dx=[x³/3+x]₀³= - 3p =27/3+3=12 - 2p

∫₀¹ 1/(f(x)−x²)² dx=∫₀¹ 1/(3x+1)² dx=(1/3)∫₀¹ (3x+1)′/(3x+1)² dx=[−1/(3(3x+1))]₀¹= - 3p =−1/12+1/3=1/4 - 2p

g(x)=3xe⁻ˣ, x∈ℝ, deci A=∫₀¹|g(x)| dx=∫₀¹3x(−e⁻ˣ)′ dx= - 2p =[−3e⁻ˣ(x+1)]₀¹=3(1−2/e) - 3p

∫₁² (f(x) − 3/x) dx = ∫₁² x³ dx = [x⁴/4]₁² = - 3p = 16/4 − 1/4 = 15/4 - 2p

G este o primitivă a funcției g ⇒ G′(x) = g(x) = (1/√x)(x³ + 3/x), x ∈ (0,+∞) - 2p (1/√x)(x³ + 3/x) > 0, pentru orice x ∈ (0,+∞), deci funcția G este crescătoare - 3p

∫₁^{√3} (1/f(x)) dx = ∫₁^{√3} x/(x⁴ + 3) dx = (1/2)∫₁^{√3} (x²)′/((x²)² + 3) dx = [(1/(2√3)) arctg(x²/√3)]₁^{√3} = - 3p = (1/(2√3))(π/3 − π/6) = π/(12√3) - 2p

2. a) ∫₁² (f(x) − 4 ln x) dx = ∫₁² 3x² dx = [x³]₁² = - 3p = 8 − 1 = 7 - 2p

2. b) ∫₁ᵉ x(f(x) − 3x²) dx = ∫₁ᵉ 4x ln x dx = ∫₁ᵉ (2x²)′ ln x dx = [2x² ln x]₁ᵉ − [x²]₁ᵉ = - 3p = 2e² − 0 − e² + 1 = e² + 1 - 2p

2. c) F′(x) = f(x) ⇒ F″(x) = f′(x), x ∈ (0, +∞) - 2p ∫₁^{√e} f(x)F″(x) dx = ∫₁^{√e} f(x)f′(x) dx = [f²(x)/2]₁^{√e} = ((3e + 2)² − 3²)/2 = ((3e − 1)(3e + 5))/2 - 3p

2. a) ∫₃⁷ f(x)/(x−1)² dx = ∫₃⁷ x dx = [x²/2]₃⁷ = - 3p = 49/2 − 9/2 = 20 - 2p

2. b) ∫₂³ x/f(x) dx = ∫₂³ (x−1)′/(x−1)² dx = [−1/(x−1)]₂³ = - 3p = −1/2 + 1 = 1/2 - 2p

2. c) ∫₀¹ x f(eˣ)/eˣ dx = ∫₀¹ x(eˣ−1)² dx = ∫₀¹ x(e²ˣ/2−2eˣ)′ dx + [x²/2]₀¹ = - 3p = [x(e²ˣ/2−2eˣ)]₀¹ − [e²ˣ/4−2eˣ]₀¹ + 1/2 = (e²−5)/4 - 2p

2. a) ∫₀² f(x)(6x² + 1) dx = ∫₀² (x + 4) dx = [x²/2 + 4x]₀² = - 3p = 2 + 8 = 10 - 2p

2. b) ∫₀² (f(x) − 4/(6x² + 1)) dx = ∫₀² x/(6x² + 1) dx = (1/12)∫₀² (6x² + 1)′/(6x² + 1) dx = [(1/12)ln(6x² + 1)]₀² = - 3p = (1/12)ln 25 = ln 5/6 - 2p

2. c) ∫₀¹ ((x + 4)/f(x)) · e^(2x) dx = ∫₀¹ (6x² + 1) · e^(2x) dx = ∫₀¹ (6x² + 1) · (e^(2x)/2)′ dx = [(6x² + 1) · (e^(2x)/2)]₀¹ − ∫₀¹ 6xe^(2x) dx = = (7e² − 1)/2 − [3xe^(2x)]₀¹ + [3e^(2x)/2]₀¹ = 2e² − 2 - 3p 2e² − 2 = m(e² − 1), de unde obținem m = 2 - 2p

2.a) ∫₀⁴ f(x)/(eˣ + 2x²) dx = ∫₀⁴ x dx = [x²/2]₀⁴ = - 3p = 16/2 − 0 = 8 - 2p

b) ∫₀¹ (f(x) − 2x³) dx = ∫₀¹ xeˣ dx = [xeˣ]₀¹ − ∫₀¹ eˣ dx = e − [eˣ]₀¹ = - 3p = e − e + 1 = 1 - 2p

c) ∫₁² (1/x) · f(x²) dx = ∫₁² (1/x) · x²(e^(x²) + 2x⁴) dx = (1/2)∫₁² (x²)′e^(x²) dx + ∫₁² 2x⁵ dx = (1/2)[e^(x²)]₁² + [x⁶/3]₁² = (e⁴ − e)/2 + 21 - 3p (e⁴ − e)/2 + 21 = (e⁴ − e)/2 + a, de unde obținem a = 21 - 2p

2.a) ∫₀¹ f(x)/eˣ dx = ∫₀¹ (3−2x) dx = (3x−2·x²/2)|₀¹ = 3p = 3−1=2 2p

b) ∫₀² f(x) dx = ∫₀² (3−2x)eˣ dx = [(3−2x)eˣ]|₀²+2[eˣ]|₀² = 3p = −e²−3+2e²−2=e²−5 2p

c) ∫ₐ¹ e³ˣ/f³(x) dx = ∫ₐ¹ 1/(3−2x)³ dx = −(1/2)∫ₐ¹ (3−2x)′/(3−2x)³ dx = (1/4)·[1/(3−2x)²]|ₐ¹ = (1/4)·(1−1/(3−2a)²), pentru orice a ∈ (−∞,1) 3p (1/4)·(1−1/(3−2a)²)=2/9 și, cum a ∈ (−∞,1), obținem a=0 2p

∫₀³ (x² + 1)f(x) dx = ∫₀³ (x² + 1 + 2) dx = [x³/3 + 3x]₀³ = - 3p. = 27/3 + 9 = 18 - 2p.

∫₁³ xf(x) dx = ∫₁³ (x + 2x/(x² + 1)) dx = [x²/2]₁³ + [ln(x² + 1)]₁³ = - 3p. = (9 − 1)/2 + ln 10 − ln 2 = 4 + ln 5 - 2p.

F este o primitivă a funcției f și f este continuă, deci, pentru orice număr real x, F(x + 1) − F(x) = ∫ₓˣ⁺¹ f(t) dt - 2p. f(t) = 1 + 2/(t² + 1) > 1, pentru orice număr real t, deci ∫ₓˣ⁺¹ f(t) dt ≥ ∫ₓˣ⁺¹ 1 dt = x + 1 − x = 1, de unde obținem că F(x + 1) ≥ F(x) + 1, pentru orice număr real x - 3p.

2.a) ∫₀² (f(x)/eˣ) dx = ∫₀² x dx = [x²/2]₀² = - 3p = 2−0 = 2 - 2p

b) ∫₋₁¹ (f(x)+eˣ) dx = ∫₋₁¹ (xeˣ+eˣ) dx = ∫₋₁¹ (xeˣ)′ dx = [xeˣ]₋₁¹ = - 3p = e+e⁻¹ = (e²+1)/e - 2p

c) f′(x) = (x+1)eˣ, deci f′(x) = 0 ⇔ x = −1 și, cum f′(x) ≤ 0, pentru orice x ∈ (−∞,−1] și f′(x) ≥ 0, pentru orice x ∈ [−1,+∞), obținem că f(x) ≥ f(−1), pentru orice număr real x - 2p f(−1) = −1/e, deci ∫_{−1−a}^{−1+a} f(x) dx ≥ −(1/e)∫_{−1−a}^{−1+a} dx = −(1/e)[x]_{−1−a}^{−1+a} = −2a/e, pentru orice a ∈ (0,+∞) - 3p

2. a) ∫₃⁵ (f(x) − √(2x² + 1)) dx = ∫₃⁵ x dx = [x²/2]₃⁵ = - 3p = 25/2 − 9/2 = 8 - 2p

2. b) ∫₁² x/(f(x) − x)² dx = ∫₁² x/(2x² + 1) dx = (1/4)∫₁² (2x² + 1)′/(2x² + 1) dx = (1/4)[ln(2x² + 1)]₁² = - 3p = (ln 9)/4 − (ln 3)/4 = (ln 3)/4 - 2p

2. c) V = π∫₀² g²(x) dx = π∫₀² (3x² + 1 + 2x√(2x² + 1)) dx = π([x³]₀² + [x]₀² + (1/2)∫₀² √(2x² + 1)·(2x² + 1)′ dx) = - 2p = π(10 + [(2x² + 1)√(2x² + 1)/3]₀²) = 56π/3, deci aπ/3 = 56π/3, de unde obținem a = 56 - 3p

2.a) ∫₀³ f(x)eˣ dx = ∫₀³ (x² + 4x) dx = [x³/3 + 2x²]₀³ = - 3p = 9 + 18 = 27 - 2p

2.b) ∫₋₁⁰ f(x)/(x + 4) dx = ∫₋₁⁰ xe⁻ˣ dx = ∫₋₁⁰ x(−e⁻ˣ)′ dx = [−xe⁻ˣ]₋₁⁰ − [e⁻ˣ]₋₁⁰ = - 3p = −e − 1 + e = −1 - 2p

2.c) G′(x) = g(x) = (x² + 4)e^(−x²), x ∈ (0, +∞), deci G″(x) = (x² + 4)′e^(−x²) + (x² + 4)(e^(−x²))′ = - 3p = 2xe^(−x²) + (x² + 4)(−2x)e^(−x²) = −2x(x² + 3)e^(−x²) ≤ 0, pentru orice x ∈ (0, +∞), deci funcția G este concavă - 2p

2.a) ∫₁² (f(x) − 4ln x) dx = ∫₁² x² dx = [x³/3]₁² = - 3p = 8/3 − 1/3 = 7/3 - 2p

b) ∫₁ᵉ ((f(x) − x²)/x) dx = ∫₁ᵉ (4ln x/x) dx = 4∫₁ᵉ ln x(ln x)′ dx = [2ln² x]₁ᵉ = - 3p = 2ln² e − 2ln² 1 = 2 - 2p

c) ∫₁² (f(x)f″(x) + (f′(x))²) dx = ∫₁² (f(x)f′(x))′ dx = [f(x)f′(x)]₁² = f(2)f′(2) − f(1)f′(1) - 3p f′(x) = 2x + 4/x, x ∈ (0,+∞); f′(1) = f′(2) = 6, deci 6(f(2) − f(1)) = m(f(2) − f(1)) și, cum f(1) ≠ f(2), obținem m = 6 - 2p

2.a) ∫₁² (f(x) − 3x ln x) dx = ∫₁² (4x + 1) dx = [2x² + x]₁² = - 3p = 8 + 2 − (2 + 1) = 7 - 2p

2.b) ∫₁ᵉ [(f(x) − 4x − 1)/x] dx = ∫₁ᵉ 3 ln x dx = ∫₁ᵉ (3x)′ ln x dx = [3x ln x]₁ᵉ − [3x]₁ᵉ = - 3p = 3e − 3e + 3 = 3 - 2p

2.c) ∫₂⁴ [(f(x) − 1)/(x² ln x)] dx = ∫₂⁴ [(4x + 3x ln x)/(x² ln x)] dx = ∫₂⁴ (4/ln x) · (ln x)′ dx + 3∫₂⁴ (1/x) dx = [4 ln(ln x)]₂⁴ + [3 ln x]₂⁴ = - 3p = 7 ln 2, deci a ln 2 = 7 ln 2, de unde obținem a = 7 - 2p

2. a) ∫₀² (f(x) − x³) dx = ∫₀² (x + 1) dx = [x²/2 + x]₀² = - 3p = 2²/2 + 2 = 4 - 2p

2. b) ∫₀¹ x²/(f(x) − x) dx = ∫₀¹ x²/(x³ + 1) dx = (1/3)∫₀¹ (x³ + 1)′/(x³ + 1) dx = [(1/3)ln(x³ + 1)]₀¹ = - 3p = (ln 2)/3 − (ln 1)/3 = (ln 2)/3 - 2p

2. c) ∫ₐ⁰ e^(−x)(f′(x) − f(x)) dx = ∫ₐ⁰ (e^(−x)f(x))′ dx = [e^(−x)f(x)]ₐ⁰ = 1 − e^(−a)(a³ + a + 1), unde a ∈ (−∞, 0) - 3p 1 − e^(−a)(a³ + a + 1) = 1 − a³e^(−a), de unde obținem a = −1, care convine - 2p

2. a) ∫₁² (f(x) + 2/(x + 1)) dx = ∫₁² 3x² dx = [x³]₁² = - 3p = 8 − 1 = 7 - 2p

2. b) ∫₁⁵ (3x² − f(x)) dx = ∫₁⁵ 2/(x + 1) dx = 2∫₁⁵ (x + 1)′/(x + 1) dx = [2 ln(x + 1)]₁⁵ = - 3p = 2 ln 6 − 2 ln 2 = 2 ln 3 - 2p

2. c) ∫₁⁴ (a + bg(x))g′(x) dx = ∫₁⁴ ag′(x) dx + ∫₁⁴ bg(x)g′(x) dx = [ag(x)]₁⁴ + [bg²(x)/2]₁⁴ = = a(g(4) − g(1)) + (b/2)(g²(4) − g²(1)), pentru orice numere reale a și b - 3p

g(1) = f(0) = −2, g(4) = f(1) = 2, deci ∫₁⁴ (a + bg(x))g′(x) dx = 4a, pentru orice numere reale a și b - 2p

2. a) ∫₁² (x² + 1)f(x) dx = ∫₁² 4x dx = [2x²]₁² = - 3p = 8 − 2 = 6 - 2p

2. b) ∫₂³ f(x) dx = ∫₂³ 4x/(x² + 1) dx = 2∫₂³ (x² + 1)′/(x² + 1) dx = [2 ln(x² + 1)]₂³ = - 3p = 2 ln 10 − 2 ln 5 = 2 ln 2 - 2p

2. c) ∫₁ᵐ (f(√x)/√x)³ dx = ∫₁ᵐ 64/(x + 1)³ dx = [−32/(x + 1)²]₁ᵐ = −32/(m + 1)² + 8 - 3p −32/(m + 1)² + 8 = 6 și, cum m ∈ (1, +∞), obținem m = 3 - 2p

2.a) ∫₁² 3(f(x) − x ln x) dx = ∫₁² 3x² dx = [x³]₁² = - 3p = 8 − 1 = 7 - 2p

b) ∫₁ᵉ (f(x)/x³) dx = ∫₁ᵉ (1/x) dx + ∫₁ᵉ (−1/x)′ ln x dx = [ln x]₁ᵉ − [(ln x)/x]₁ᵉ + ∫₁ᵉ (1/x²) dx = - 3p = 1 − 1/e − [1/x]₁ᵉ = 2 − 2/e = 2(1 − 1/e) - 2p

c) g(x) = (x + 1)/(x² + x ln x) ≥ 0, pentru orice x ∈ [1, e], deci 𝒜 = ∫₁ᵉ |g(x)| dx = ∫₁ᵉ (x + 1)/(x² + x ln x) dx = - 2p = ∫₁ᵉ (1 + 1/x) · 1/(x + ln x) dx = ∫₁ᵉ (x + ln x)′ · 1/(x + ln x) dx = [ln(x + ln x)]₁ᵉ = ln(e + 1) > ln e = 1 - 3p

2. a) ∫₃⁵ (f(x) − x³) dx = ∫₃⁵ x dx = (x²/2)|₃⁵ = - 3p = 25/2 − 9/2 = 8 - 2p

2. b) ∫₀² x²/(f(x) − x + 2) dx = ∫₀² x²/(x³ + 2) dx = (1/3)∫₀² (x³ + 2)′/(x³ + 2) dx = (1/3)ln(x³ + 2)|₀² = - 3p = (ln 10)/3 − (ln 2)/3 = (ln 5)/3 - 2p

2. c) g(x) = (x² + 1)e⁻ˣ și G′(x) = g(x), pentru orice x ∈ (0,+∞), deci G″(x) = (−x² + 2x − 1)e⁻ˣ - 3p G″(x) = −(x − 1)²e⁻ˣ ≤ 0, pentru orice x ∈ (0,+∞), deci funcția G este concavă - 2p

2. a) ∫₁ᵉ x²(f(x) + (2ln x)/x³) dx = ∫₁ᵉ (1/x) dx = [ln x]₁ᵉ = - 3p = ln e − ln 1 = 1 - 2p

2. b) ∫₁^{√5} x·f(x² + 3) dx = (1/2)∫₁^{√5} (x² + 3)′·f(x² + 3) dx = (1/2)[F(x² + 3)]₁^{√5} = - 3p = (1/2)((ln 8)/64 − (ln 4)/16) = −(5ln 2)/128 - 2p

2. c) ∫ₑ^{e²} x·F(x) dx = ∫ₑ^{e²} (ln x)/x dx = [(ln² x)/2]ₑ^{e²} = (ln²(e²) − ln² e)/2 = 3/2 - 3p (a² − 1)/2 = 3/2, de unde obținem a = −2 sau a = 2 - 2p

Reguli generale:

2. a) ∫₀¹ eˣf(x) dx = ∫₀¹ (eˣ + 2x) dx = [eˣ + x²]₀¹ = - 3p = e + 1 − 1 = e - 2p

Reguli generale:

2. b) ∫₋₁⁰ f(x) dx = ∫₋₁⁰ (1 + 2xe⁻ˣ) dx = [x − 2(x + 1)e⁻ˣ]₋₁⁰ = - 3p = −2 − (−1) = −1 - 2p

Reguli generale:

2. c) F′(x) = f(x), pentru orice x ∈ ℝ, deci ∫₀¹ F(x)f″(x) dx = [F(x)f′(x)]₀¹ − [f²(x)/2]₀¹ = = F(1)f′(1) − F(0)f′(0) − (f²(1) − f²(0))/2 - 3p f′(x) = 2(1 − x)e⁻ˣ, deci f′(1) = 0 și, cum F(0) = 0, obținem ∫₀¹ F(x)f″(x) dx = −2(e + 1)/e², deci a = −2 - 2p

Reguli generale:

2. a)

∫₀¹ (4 − f²(x)) dx = ∫₀¹ (4 − 4x²/(x² + 1)) dx = 4∫₀¹ 1/(x² + 1) dx = 4 arctg x│₀¹ = - 3p

= 4(arctg 1 − arctg 0) = π - 2p

Reguli generale:

b) 𝒜 = ∫₀¹ |f(x)| dx = ∫₀¹ 2x/√(x² + 1) dx = ∫₀¹ (x² + 1)′/√(x² + 1) dx = - 3p

= 2√(x² + 1)│₀¹ = 2(√2 − 1) - 2p

Reguli generale:

c) ∫₁² f(x²)/x dx = ∫₁² 2x²/(x√(x⁴ + 1)) dx = ∫₁² 2x/√(x⁴ + 1) dx = ∫₁² (x²)′/√((x²)² + 1) dx = - 3p

= ln(x² + √(x⁴ + 1))│₁² = ln((4 + √17)/(1 + √2)) = ln(√34 − √17 + 4√2 − 4) - 2p

Reguli generale:

2. a) ∫₁³ (f(x) + 2 ln x) dx = ∫₁³ (2x + 3 − 2 ln x + 2 ln x) dx = ∫₁³ (2x + 3) dx = [x² + 3x]₁³ = - 3p = 9 + 9 − 1 − 3 = 14 - 2p

Reguli generale:

2. b) ∫₁ᵉ (2x + 3 − f(x)) dx = 2∫₁ᵉ ln x dx = [2x ln x]₁ᵉ − 2∫₁ᵉ x · (1/x) dx = - 3p = 2e − 2(e − 1) = 2 - 2p

Reguli generale:

2. c) ∫₀¹ x² f(x³ + 1) dx = (1/3)∫₁² f(t) dt = (1/3)(∫₁² (2t + 3) dt − 2∫₁² ln t dt) = - 3p = (1/3)[t² + 3t − 2t ln t + 2t]₁² = (1/3)(4 + 6 − 4 ln 2 + 4 − 1 − 3 − 2) = 4(2 − ln 2)/3 - 2p

2.a) ∫₀¹ f(x)√(x² + 2) dx = ∫₀¹ (x² + 2 + x) dx = [x³/3 + 2x + x²/2]₀¹ = - 3p = 1/3 + 2 + 1/2 = 17/6 - 2p

b) F′(x) = f(x) = (x² + x + 2)/√(x² + 2), unde F: ℝ → ℝ este o primitivă a funcției f - 3p x² + x + 2 > 0, pentru orice număr real x, deci F′(x) > 0, pentru orice număr real x, de unde obținem că F este crescătoare pe ℝ - 2p

∫₀¹ g(x) dx = ∫₀¹ x′√(x² + 2) dx = [x√(x² + 2)]₀¹ − ∫₀¹ x · x/√(x² + 2) dx = √3 − ∫₀¹ (√(x² + 2) − 2/√(x² + 2)) dx, deci ∫₀¹ g(x) dx = [(1/2)(√3 + 2ln(x + √(x² + 2)))]₀¹ = √3/2 + ln((1 + √3)/√2) - 3p √3/2 + ln((1 + √3)/√2) = √3/2 + ln((a + √3)/√2), de unde obținem a = 1, care convine - 2p