Esercizi di Analisi matematica 1

∫(xⁿ+1)(xⁿ⁻¹)dx

∫√(x⁴−x)dx

∫(tan(x))(sec⁴(x))dx

∫(2+x)/[(x²+4)(x²+1)]dx

∫4/(x²−5^½)dx = 2∫dx/√(x⁴−x) + x/√(x⁴−x) + C

∫√(2x-ln(x))dx

∫[1/(x²−5x+6)]dx = ∫[1/(x-2)]−[1/(x-3)]dx

x³+x⁺²ⁿ

∫x⁻²(3x+1+x)dx =

∫[√x+4]/[x²(x+8)]dx

x=x±^½

∫3log√x = ∫ 3log√[x-1]dx + C

∫(exm+dx

(f(x)=ln(x), p(x)=ex)

x(e^ln(x))dx

(x+t²)/4−√3cos(x)dx

e^xsen(3x)dx

f(x)=e^x g(n)=sen(3x)

f'(x)=e^x g'(x)=cos(3x)⋅3

= e^x⋅sen(3x)−∫e^xcos(3x)⋅3dx

f(n)=e^xg(x)=−cos(3x)

= e^x⋅sen(3x)−3[e^xcos(3x)−∫e^xsen(3x)⋅3]dx

f(n)=e^x g(x)=cos(3x)

= e^x⋅cos(3x)−3∫e^xsen(3x)dx

∫e^xsen(3x)dx =sen(3x)⋅3

f(n)=e^xsin(3x)−e^xcos(3x)−∫e^xsen(3x)dx

f(x)=arctom(cx) * n

p(n):=arctom(cx)=∫arctom(cx)*n dx

∫^c_aarctom(x)→∫^b_aarctom(x)=n2

+arctom(x)⋅n2+∫arctom(x)⋅1dx+∫arctom(x)∗1dx

=n2arctom(x)−n(e)n2arctom(x)+C

log(n)(1+sen(cx))+C.

1+e^xdx

1ⁿ+t+e^x++1ⁿ+t−e^xdx

∫=1 to e^x(13)

ⁿ_n−t^t−1

∫1 to e^xdx

arctom(e^x)+e^xlog(e^x+C)+C

arctom(e^x)+

1+log(e^x) + C

= t: e^u(x)

1+log(e^xt)+C

e^u(x) − ln(21e^u(x)) + t +C.

g(u)=∫

1⁴e + dt

e^u

1⁴

u = √1+√1+4 + C.

x + arctom

log(1+sen(x))+c

x/sinx cosx

tg(x)·cot(x).

ds = ∫sen(x)dx

= sin(x) = (sen(n)−15

u=∫1+u⋅u

dx =

f(x) = x³ - 3x + 2

DOMINIO: x ∈ R

SHAME TEMA

f(x) = (x³ - 3x) + 2 ⇒ x = x³ + 3x + 2

INTERSEZIONI:

asse x

y = 0

x = 3x² + 0 = 20

x - 2 = x = -2

y = 0

(-2,0)

(1,0)

SEZIONE F(x) ≥ 0

x³ - 3x + 2 > 0

(x + 2) (x - 1) (x - 1) > 0

lim x → ∞

f(x)

x³ - 3x + →

x → -∞

diinci!

f'(x) ≥ 0

⇒ 3x² ≥ 3x

x > 0

f'(x) ≥ 6x

∀ x > 0

(x + 2) (x - 1) (x - 1)

(-2) -2 = +4

(1, 0) 0 = 0

f(x) = e^-1/x3 / x+4

Dominio:

x+3 ≠ 0 → x ≠ -3 x+4 ≠ 0 → x ≠ -4 D: ∀x ∈ ℝ / {-3, -4} ∀x ∈ (-∞, -4) ∪ (-4, -3) ∪ (-3, +∞)

Intersez:

y=0 → e^-1/x3 = 0 → 2^1/x3 = 0 x=0 → 0, 3√(2/4)

Asintoti:

lim_x→-3⁺ e^-1/x3 / x+4 = lim_x→-3⁻ e^-1/x3 / x+4 = 0 lim_x→-4⁺ e^-1/x3 / x+4 = lim_x→-4⁻ e^-1/x3 / x+4 = 0 lim_x→-∞ e^-1/x3 / x+4 = lim_x→+∞ e^-1/x3 / x+4 = 0 lim_x→-4 √(e^-1/x3) / (x+3) = 1

f'(x) = [e^-1/x3(x+4) - e^-1/x3(1)] / (x+3)² = e^-1/x3 / (x+3)²

f'(x) > 0 → (x+1)² > 0 ∀x ∈ D e^-1/x3 > ∀x ∈ ℝ

x+4 > 0 per x ≠ -4 x+3 > 0 ∀x ∈ D

x² + 8x +16

Conclusions:

f(x)= x² - log(x)

Polinomio: x > 0 => x ∈ (0,+∞)

Asintoto: f(x) => e^x2-log(x)log x

Segno

f(x) => x > 0 f(x) => mai

f(x) => x - log(x)

x² + log x

x / x² + 1 > 0 ∈ Df(x)= x > 0

∫ tan x sec²xdx

= ∫ tan²x + tanx sec xdx

∫ (x² + 1)^-2 = cos x

2. ∫ dx

3. ∫

x²-1

dv =

4. ∫ x /(x+1)

u =

arctan u + c

f(x)=2x => f’(x)= x

g’(x)= xe^x

f(x)= f’(x) = x