Analisi matematica 1

∫_c^x √³x + 4 dx = e^x

∫ e^x(e^{½ x})⁴ dx = e^-x-a

e^c dx = b^e√ e^x-a

= ¹/_x⁴(e^½ x)^½ + c = (e^½ 3) (¹/₄) + k

anal=log(c+1/x)

parametr per P (0; 1)

= tg √ dy =

= ∫ sen y _dy/^cos x

= ) sen y ⋅ ₁/^cos x dx

∫cp (cos x)

= - ¹/_dx (a(cos x) - x) = - log |_cosx

=^x/_{x⁷ dx}

= ∫ x/(e^x - 1.1 e^{c x}) dx

= ∫ ^x/_e = sen x

∫ (x - π_/2) dx

= 5 (x^t/g) L(x)

= ∫ sen √x/√x

= ∫ - e²cos √ π dx

= D(cos x)₁/² sen x ⋅ dx

D (cos √x) = - sen +

D (∫ F ) = ∧ ⅔

∫ ∫ ^x/_{x² - a dx}

= D (x²-a) - 2x

=¹/_z= cop log |^{x +1}| - πc

= ¹/_{X a}² dx

D (arcp x) = - ¹/_{4 + (X^X)/(4)}

= ¹/₄ [^x/(ξ x)]

dy = D(¹/_z) = ¹/₂

guarda slide

∫ 1 + X⁸

(^Y/_⋅e⁴) dy

D(x-e) ¹/₄ - (^-1)e^-x = ^-1 + e^-a1

-log |X e²｜1 ∫ X

y’=-e^-2+x

y’≥e^-x

x e^x dx

D ( x^x ) = e^x derivata esponente

= { ∫x e x dx = ∫e^x +kc

= { ¹/_{√x cos (√x)} dx

= ¹/_{cos +} = D ( tg +1 )

D ( F(x) ) = ¹/_2√x

= ¹/_√2√x dx = 2 tg √x +k

= ∫ ¹/_{x(1+ log x)²} dx

= [ ¹/_x ∣ ¹/_MI/log²xdx ]

= D(log x) ¹/_1+x = … -D (arc tg+m)

- arcsen ( log x ) +xz

= ∫ ^ex/_√1-e^2x dv

= D (arcsen) = ¹/_√1-y

= e^x = D(e^x)

= ∫ e^{x 1}/_√1-I²(x) dx = arcsen (e^x)

∣ ^cos5xdy

= [cos x ∙ cos⁴ x dx =

= ∫[ cos x ( 4 - sen²x)²dx = - sen x < cos⁴x

=∫[ 4+sen²x - sen²x ) ∙ d(sen x)

cos x dx = D (sen x)

[| (1+t²-1)^dt ]

= ( devendo sudr

D ( f(x) ∙ g(x) ) - l' (x) ∙ d ( x ) + s ( x ) ∙ I ( x )

(f(x) ∙ x)( x ) = ∫(l(x) ∙ 1(x) +

f(x) ∙ I(x) ) dv

∫ l’ (x) ∙ 1(x)) dv = ∫ ( l(x) ∙ f(x) )

- ∫f ( x ) ∙ 1’ (x)) dv

lo domo denore …

¹/_{3 e³ x²} dx =

funzione on vevnore

= ep - | x ∙ x²-3 ∣3x² ∙ e^xdx

= 3 { ∫e^-xx dx = 3 ∫ x^-xe^-x dx

{ 2n e^-xdx ∫

2 ∫ x e^x dx = 2 ∫ {xe^x - ∫ 1 e^x dx} = 2 ∫ {xe^x - e^x}dx

b(x)

= x³ e^-3 - 3 ∫ x² e^-2 - xe^-xy} + x 3e^-3 x e^-6 + 6 x e^-x - 6e e^-xe

⋅arcϕ ^π/_{4 4} dx

∫ arcϕ ∂ = ∫ x ⋅ e^-x dx

D (e^-x) = -e^-x

-xe^-x + ∫∫ -x ⋅ e^-x dx = - x ⋅ e^-x - e^-x

(1) ∫ ^{log 2y}/_x² dx

D (x^1/2) = 1/_{2 x^1/2}

2 ∫ ¹/_√x ⋅ log x ⋅ dx = 2 √x log x dx

bigop(x)

a ⋅ {∫ √x log x - ∫ √x⁴

∫ e^x sen x ⋅ dx = e^x D(sen x)

D(sen x) = cos x

Questo equivale a

cos ⋅ dx ∧ e e^{n ∼ en x} dx + c

25/11/2021

Calcolare i seguenti integrali

1. ∫ 3x^2 - 8x^2 - 6x + 8 dxb = -6 c = 8b² - 4acx^2 - 6x + 8 = 0 => D => 6 ± 2 2 \ 4=> x^2 - 6x + 8x^2 - 6x + 8 => (x-2)(x-4)

Scompo: 3x - 4 A B

.........................................................

Dunque

∫ 3x - 4 = _A + ∫ dx............. x - 4 x-2

2. ∫ 5x + 2 x^2 + 2x - 3 dxb² - 4acx^2 + 2x - 3 = 0x= -2 ± 0 ........... 3 \ 1 -1=> x^2 + 2x - 3 = (x-1)(x+3)

Scompo: 5x + 2 = A + B............(x-1)(x-3) x - 1 x + 3

.........................................................

Dunque

∫ 5x + 2............(x-1)(x-3) dx = ∫ dx - 1/2 log|x-1| 3/4 ∫ 1 ....... x-1 + 3/4 + 3/2 x-1 13/4 log |x+3| + c

3. ∫ x + 2 ........ x^2 + 2x + 5b = 4 - 3 c = -5b² < acScompo: x^2 - 2x + 5 = x^2 - 2 + (....................................=> 1/2 ∫ x(x-2) + a 5 dx

Integratore per sostituzione

Calcolare i seguenti integrali effettuando le opportune sostituzioni

1. \(\int \frac{x}{\sqrt{1+x^2}}\log (x)\ dx\)

   sost: \(t=\log x \rightarrow e^t=x \rightarrow dx=e^t dt\)

   Quindi \(\int \frac{e^t}{\sqrt{1+e^{2t}}} \cdot e^tdt = \int \frac{dt}{\sqrt{1+t^2}}\)

   \(= \frac{1}{\sqrt{2}} \cdot \arctan \left(s\right)+c = \frac{1}{\sqrt{2}} \cdot \arctan \left(\frac{\log x}{\sqrt{2}}\right)+c\)

2. \(\int \frac{x^5}{\sqrt{1-x^2}} dx = \int \frac{x^3 \cdot x^2}{\sqrt{1-x^2}} dx\)

   sost: \(t=1-x^2 \rightarrow dx = -2x\cdot dx\)

   Quindi \(\int x^3\cdot dx = -\frac{1}{2} \int \frac{1-t}{\sqrt{t}}dt = \frac{1}{2} \left(-\frac{1}{3} t \sqrt{t} + \sqrt{t}\right)\)

   \(=\int (1-x^2)^\frac{1}{2} = \frac{1}{4} - \frac{1}{4}\cdot\sqrt{4-x^2}+c\)

3. \(\int \frac{x^5}{\sqrt{x^3-4}} dx\)

   Quando \(\int \frac{x^5}{x^3-4}\ dx = \frac{1}{3} \int \frac{t^{4/3}}{t} dt\)

   \(= \frac{3}{5} \int \frac{1}{u} dt = \frac{1}{3} \left(-\frac{1}{3}\cdot\sqrt{5t}\right)\)

   \(= \frac{1}{9} \left(\sqrt{5t}-t^{11/2}\right)+c\)

4. \(\int \frac{dx}{x^2 \sqrt{1+x^2}}\)

   sost: \(t = \frac{1}{x} \rightarrow dt = -\frac{dx}{x}\)

   \(= -\int \frac{tx\ dt}{\sqrt{1+t^2}} = -\int \frac{t}{\sqrt{t^4}}\ dx\)

   \(=\frac{1}{2} \int \frac{ds}{\sqrt{15}} = \sqrt{15}+c = -\sqrt{2048t}\)₂+c

5. \(\int \sqrt{2\cdot4-1}\ dx\)

   Prova la sostituzione \(e^x\rightarrow e^xt\nabla\frac{e}{x-\sigma}\)

   \(\int =5\sqrt{\nu+1}\ dx\)

prova la sostituzione

prova con

quando

a)

b)

quindi

c)

Dunque

Quindi

S_W/3 = ∫_W/3¹      dx = ∫₀¹      dt

      2cosπx+cosπx - 1      2t(2 - t)

= ∫₀^W/3      dt =      -      - (    - )

      1/(2-t)      1/(1-t)     1/(2-t)     1/t)

= -¹/₂ ∫₀¹   (     ) dt

     (   - 1)    /t

= -¹/₂ [  log H+2] + ¹/₄ [  log+]^W/3

= ¹/₂ log(2 -  ) - ¹/₂ log  

25/11/2022

Calcolare i seguenti integrali