Integrali Propri da esame

Integrali propri

1. ∫₀¹ ln(arctan x + 1) / (x² + 1)(arctan x + 1)² dx
2. ∫₁² 1/x³ arctan(1/x) dx
3. ∫ √(3x+6) / (2x²-3x+3) dx
4. ∫(3x+1) / (x³+2x²+x+2) dx
5. ∫(2x³+x²+4) / (x²+4)² dx
6. ∫ cos x / (2-sen²x) dx

Integrali propri

1. ∫₀¹ ln(arcotan x + 1) / (x² + 1)(arcotan x + 1)² dx
2. ∫₁² 1 / x³ arcotan 1 / x dx
3. ∫ (3x + 6) / 2x²-3x+3 dx
4. ∫ (3x + 1) / x³ + 2x² + x + 2 dx
5. ∫ (2x² + x² + 4) / (x² + 4)² dx
6. ∫ cos x / 2 - sen²x dx

Integrali propri

1. ∫₀¹ ln () ∫ dx
2. ∫₂^x arctan ∫ dx
3. ∫ √
4. ∫
5. ∫
6. ∫
7. ∫
8. ∫
9. ∫
10. ∫ dx (not.)
11. ∫ arctan ∫ dx (not.)
12. ∫
13. ∫
14. ∫
15. ∫

1. ∫₀¹ ∫ dx t = dt = ∫ ∫ => ∫ -t . t dt => = - ∫ [- -1 - [-1]- - -1 = [-[- [-]]]] = -[[[-[+1 = -4[ -+]]]

∫₁^{2   1}⁄_x³ arctan( ¹⁄_x ) dx

Per parti -¹⁄_2x²   arctan( ¹⁄_x ) -  ∫-¹⁄_{x² + 1}   ( -¹⁄_2x²) dx

-¹⁄_2x² arctan( ¹⁄_x ) - ¹⁄₂   ∫ ¹⁄_{x²(x² + 1)} dx

¹⁄_{x²(x² + 1)} => ^{Ax + B}⁄_x² + ^{Cx + D}⁄_{x² + 1} => ^{Cx³ + Dx² + Ax³ + Bx}⁄_{x²(x² + 1)}(C + A = 0D + B = 0A = 0B = 1x C = 0x D = -1A = 0B = 1)=  ¹⁄₂[∫¹⁄_x² dx - ∫¹⁄_x²+1 dx]

-¹⁄_2x²   aretg ( ¹⁄_x ) - ¹⁄₂ [ -¹⁄_x - aretg x ]​x x ​ ²​[ -¹⁄_2x²   arctan (¹⁄_x) - ¹⁄₂ ( -¹⁄_x - aretg x )]₁[ -¹⁄_4x²   arctan   (¹⁄₂) - ¹⁄₂ (-¹⁄₂ - arctan   2 ) ] - [ -π/^{8 x²} - ¹⁄₂ ( -π/₄) ]

3

∫ ^3x+6⁄_x²-3x+2 dx

x²-3x+2 → (x-1)(x-2)

3x+6 ⁄ x²-3x+2 → ^A⁄_x-1 + ^B⁄_x-2 → ^Ax-2A+Bx-B⁄_(x-1)(x-2) → ^x(A+B)-2A-B⁄_(x-1)(x-2){A + B = 3-2A - B = 6A = 3 - B-2(3-B) - B = 6B = 12-∫⁹⁄_(x-1) dx + ∫¹²⁄_(x-2) dx

-9 ln |x-1| + 12 ln |x-2| + e

4

∫ ^3x+1⁄_x³+2x²+x+2 dx

x³+2x²+x+2 = (x²+1)(x+2)

3x+1 ⁄ x³+2x²+x+2 → ^A⁄_(x+2) + ^Bx+C⁄_(x²+1) = ^{Ax2+Ax+Bx2+Cx+2C}⁄_(x+2)(x²+1) {A = -1B = 1C = 1∫ ¹⁄_x+2 dx + ∫^x+1⁄_x²+1 dx → -ln |x+2| +¹⁄₂ ln |x²+1| + arctg x + e

∫(2x³ + x² + 9) / (x² + 4)² dx

2x³ + x² + 9 = (Ax + B)(x² + 4) + (Cx + D)

Ax + B / (x² + 4) + Cx + D / (x² + 4)²

(Cx + D) + (Ax + B)(x² + 4) / (x² + 4)²

Δx³ + ΓAx + Bx² + 4B + Cx + D / (x² + 4)² = x³ + x² + 4x + 4B

A=2B=-1C=-8D=0

∫(2x + 1) / (x² + 4) dx + ∫-8x) / (x² + 4)² dx

Separati ∫(2x + 1) / (x² + 4) dx = ln |x² + 4|

∫1 / (x² + 4) = 1/2 arctan (x/2)

∫-1 / (x² + 4) dx = {∫1 / x² + a² dx = 1/2 arctan (x/a)}

t=x²+4dt=2x dxdx=1/2x dt-8∫ x / (x² + 4)² dx-8∫ * / t²-8/2 ∫1 / t² dt-4∫1 / t² dt -> -4 * -1/t -> 4/t -> 4 / (x² + 4)

ln |x² + 4| + 1/2 arctan (x/2) + 4 / x² + 4 + C

6

∫(cos x / 2 - sin²x dx

ε = sin x dε = cos x dx

∫1 / 2 - t² dt -> ∫-1 / (-2 + t²) dt

∫-1 / t² - 2 dt ecriture come potenze - ∫1 / t² + 1/2 dt

∫1 / x² - a² dx = 1/2α ln |x - a/x + a| + C => 1/2√2 ln |ζ - √2/ζ + √2| + C-ln |sen x - √2| - ln |sen x + √2| + C/2 ₀ √2

7

∫ ^{e2x - 1}/_{2e^2x - 1} dx

t = e^2x2x = ln tx = ln t/2dx = 1/2 . 1/t dt

(t - 1)2t - 11/2∫ (t - 1)/t(2t - 1) dt

A = 1B = -11/2 [ ∫ 1/t dt + ∫ -1/2t dt-> 1/2 [ ln |t| - 1/2 ln |2t - 1| ]

1/2 ln |e^2x - 1| - 1/4 ln |2e^2x - 1| + t_e

8

∫ 1/_{√x + √x̄} dx ⇒ ∫ 6t⁵/ _{t³ + t²} dt

t = √3√xx = t⁶dt = 6t⁵ dt

t⁵t³ + t²-t²-t²/_{t³ + t²}6 [t² + 1 +∫ t⁵ dt -> t² - t + 1 +-1∫ t dt + ∫ - t² ∫ t³/3 - t²/2 + t - ln |t + 1| ]- ∫ 1/_{t + 1}

6 [√x/3 - √x/2 + 6√x - ln |√x + 1| 2√x - 3√³x + 6√x - 6 ln |√x + 1| + t_e

\(\int \frac{2\sin^2x + 3 \sin x + 3}{\sin x - 1}(\text{num} \, e^{x} + 1)\) \(\cos x \, dx\)

\(t = \text{num} \, x\)

\(dt = \cos x \, dx\)

\(\int \frac{2t^2 + 3t + 3}{(t - 1)(t^2 + 3)} \, dt\)

\(\frac{2t^2 + 3t + 3}{(t - 1)(t^2 + 3)} \rightarrow \frac{A}{t - 1} + \frac{Bt + C}{t^2 + 3}\)

\(At^2 + 3A + Bt^2 + Ct + (t - 1)(t^2 + 3)\) \(\Rightarrow \frac{t^2(A + 0) + t(C - B) + 3A - C}{(t - 1)(t^2 + 3)}\)

A + B = 2C - B = 33A - C = 3A = 2B = 0C = 3

\(\int \frac{2}{t - 1} \, dt + \int \frac{3}{t^2 + 3} \, dt \Rightarrow 2 \int \frac{1}{t - 1} \, dt + 3 \int \frac{1}{t^2 + 3} \, dt\)

\(2 \ln |t - 1| + 3 \left[ \frac{1}{t^2 + 3} \, dt \right]\)

\(2 \ln|t - 1| + \frac{\sqrt{3}}{3} \, \arctan \left(\frac{t}{\sqrt{3}}\right) + e\)

\(2 \ln |\sin x - 1| + \sqrt{3} \arctan \left(\frac{\sin x}{\sqrt{3}}\right) + e\)

\(\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \arctan \left(\frac{x}{a}\right)\)

10

\(\int \frac{x + 14}{x^2 + 2x + 5} \, dx\)

\(\Rightarrow \int \frac{x + 14 + 1 - 1}{x^2 + 2x + 5} \, dx\)

\(\Rightarrow \frac{x + 1}{x^2 + 2x + 5} \, dx + \left( \frac{13}{x^2 + 2x + 5} \right)\)

\(\frac{2(x + 1)}{x^2 + 2x + 5} \, dx + 13 \int \frac{1}{(x + 1)^2 + 4} \, dx\)

\(= \ln |x^2 + 2x + 5|\)

\(13 \int \frac{1}{(x + 1)^2 + 4} \, dx\)

\(= 13 \int \frac{1}{t^2 + 4} \, dt\)

\(13 \cdot \frac{1}{2} \arctan \left(\frac{t}{2}\right)\)

\(\ln |x^2 + 2x + 5| + \frac{13}{2} \arctan \left(\frac{x + 1}{2}\right) + C\)

\(\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \arctan \left(\frac{x}{a}\right)\)

11

∫ \arctan\left(\frac{1}{x}\right) dx

∫ \arctan\left(\frac{1}{t}\right) dt

t = x →dt = dx

Per parti t \cdot \arctan\left(\frac{1}{t}\right) - ∫ \frac{t}{1+\left(\frac{1}{t}\right)^{2}}\left(-\frac{1}{t^{2}}\right) dt

t \cdot \arctan\left(\frac{1}{t}\right) - ∫ \frac{t}{t^{2}+1} dt

t \cdot \arctan\left(\frac{1}{t}\right)\left(\frac{1}{2}\right) ∫ \frac{2t}{t^{2}+1} dt → t \cdot \arctan\left(\frac{1}{t}\right) + \frac{1}{2} \ln|t^{2}+1| + c

t² = x² - 2x + 1× \arctan\left(\frac{1}{x-1}\right) - \arctan\left(-\frac{1}{x-1}\right) + \frac{1}{2} \ln|x^{2} - 2x + 1| + c

12

∫ \frac{\sqrt{x^{2}+1}}{x} dx

∫ \frac{t}{\sqrt{t^{2}-1}} \frac{t}{\sqrt{t^{2}-1}} dt

∫ \frac{t^{2}}{t^{2}-1} dt

∫ \frac{t^{2}-1+1}{t^{2}-1} dt

∫ \frac{t^{2}-1}{t^{2}-1} dt + ∫ \frac{1}{t^{2}-1} dt

− ∫ \frac{1}{x^{2} - a^{2}} dx = \frac{1}{2a} \ln\left|\frac{x-a}{x+a}\right| + c

\sqrt{x^{2}+1} + \frac{1}{2} \ln\left|\frac{\sqrt{x^{2}+1}-1}{\sqrt{x^{2}+1}+1}\right| + c

∫ ^{1 - ex}/_{e^2x + 1} dx

- ∫ ^ex/_{e^2x + 1} dx + ∫ ¹/_{e^2x + 1} dx

- ∫ ^ex/_{e^2x + 1} dx + ∫ ¹/_{e^2x + 1} dx

- arctan (e^x) + ∫ ¹/_{e^2x + 1} dx

1/2 ∫ ¹/_t 1/_t dt -> 1/2 ∫ ¹/_t(t+1) dt

t = e^2x2x = ln tx = ^{ln t}/₂dx = 1/2 ∙ 1/_t dt

1/_t(t+1) -> ^A/_t + ^B/_t+1

^{A t + A + B t}/_{t (t + 1)} = ^{t (A + B) + A}/_{t (t + 1)}{ A + B = 0 {A = 1 A = 1B = -1

1/2 [ ∫ ¹/_t dt - ∫ ¹/_t+1 dt ]

1/2 [ ln |t| - ln |t + 1| ]

- arctan (e^x) + 1/2 (ln |e^2x| - ln |e^2x + 1|) + C

13

∫ x⁵ e^x2 dx

1/2 ∫ e^t t² dx

Parti 1/2 [ e^t t² - 2 ∫ e^t t o 1 dt ] -> Parti 1/2 [ e^t t² - 2 (e^t t - ∫ e^t) ]

1/2 [ e^t t² - 2 e^t t + 2e^t]

1/2 e^t (t² - 2t + 2) + e

1/2 e^x2 (x⁴ - 2x² + 2) + e

14

∫ x² lu x dx

x³/3 lux -1/3 ∫ x³ . 1/x dx

x³/3 lux - 1/3 x³/3

x³/3 lux - x³/9 + e

f: t²        f': 2t g': e^t     β = e^t

f: t        f': 1 g': e^t       β = e^t

f: lux     f': 1/xg': x²     β = x³/3

Varianti 7

^{∫ e2x - 1}/_{e^2x - 4} dx

∫ e^2x/_{e^2x - 4} dx - ∫ 1/_{e^2x - 4} dx

1/2 ∫ 1/_t dt

1/2 ln|t| : = 1/2 ln |e^2x - 4|

∫ 1/_{e^2x - 4} dx => 1/2 ∫ 1/_t(t+4) dt

A/t + B/_t+4 =>Bt + At + 4A/_t(t+4){A + B = 0 {A = 3/4{4A = 1 {B = -1/4

1/2 (1/4 ∫ 1/_t dt - 1/5 ∫ 1/_t+4 )

1/2 (1/4 ln|t| - 1/4 ln|(t+4)|) = 1/2 (1/4 ln |e^2x - 4| - 1/4 ln |e^2x|)

1/2 ln|e^2x - 4| - 1/2 ( 1/4 ln |e^2x - 4| - 1/4 ln |e^2x - 1| ) + e