Analisi matematica 1

25/11/2022 Calcolare i seguenti integrali

∫_c^x √³x + 4 dx = e^x∫ e^x(e^{½ x})⁴ dx = e^-x-ae^c dx = b^e√ e^x-a= ¹/_x⁴(e^½ x)^½ + c = (e^½ 3) (¹/₄) + kanal=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 ⋅ dxD (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 dxD (arcp x) = - ¹/_{4 + (X^X)/(4)}= ¹/₄ [^x/(ξ x)]dy = D(¹/_z) = ¹/₂guarda slide∫ 1 + X⁸(^Y/_⋅e⁴) dyD(x-e) ¹/₄ - (^-1)e^-x = ^-1 + e^-a1-log |X e²｜1 ∫ Xy’=-e^-2+xy’≥e^-xx e^x dxD ( 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 4x=∫[ 4+sen²x - sen²x ) ∙ d(sen x) cos x dx = D (sen x)[| (1+t²-1)^dt ]= ( devendo sudrD ( 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}dxb(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 dxD (e^-x) = -e^-x-xe^-x + ∫∫ -x ⋅ e^-x dx = - x ⋅ e^-x - e^-x(1) ∫ ^{log 2y}/_x² dxD (x^1/2) = 1/_{2 x^1/2}2 ∫ ¹/_√x ⋅ log x ⋅ dx = 2 √x log x dxbigop(x)a ⋅ {∫ √x log x - ∫ √x⁴∫ e^x sen x ⋅ dx = e^x D(sen x)D(sen x) = cos xQuesto equivale acos ⋅ 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 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 sostituzioneprova conquandoa)b)quindic)DunqueQuindiS_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  

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