Pregunta 1. ∫ x/(x 2 + 3x + 2) dx
Solución:
Dado que I = ∫ x/(x 2 + 3x + 2) dx
Sea x = md/dx (x 2 + 3x + 2) + n
= m(2x + 3) + norte
x = (2m)x + (3λ + n)
Al comparar los coeficientes de x,
2m = 1
metro = 1/2
3m + norte = 0
3(1/2) + norte = 0
n = -3/2
yo = ∫(1/2(2x + 3) – 3/2)/(x 2 + 3x + 2) dx
= 1/2 ∫(2x + 3)/(x 2 + 3x + 2) dx – 3/2 ∫1/(x 2 + 3x + 2) dx
= 1/2 ∫(2x + 3)/(x 2 + 3x + 2) dx – 3/2 ∫1/(x 2 + 2x(3/2) + (3/2) 2 – (3/2) 2 + 2) dia
= 1/2 ∫(2x + 3)/(x 2 + 3x + 2) dx – 3/2 ∫1/((x + 3/2) 2 –(1/2) 2 ) dx
= 1/2log|x 2 + 3x + 2| – (3/2) × (1/2 × (1/2))log|(x + 3/2 – 1/2)/(x + 3/2 + 1/2)| +c
Como sabemos que ∫1/(a 2 – x 2 )dx = 1/2a log|(x – a)/(x + a)| +c]
Por lo tanto, yo = 1/2 log|x 2 + 3x + 2| – (3/2) × (1/2 × (1/2))log|(x + 1)/(x + 2)| +c
Pregunta 2. ∫(x + 1)/(x 2 + x + 3) dx
Solución:
Dado que I = ∫(x + 1)/(x 2 + x + 3) dx
Consideremos x + 1 = md/dx(x 2 + x + 3) + n
x + 1 = m(2x + 1) + norte
x + 1 = (2m)x + (m + n)
Al comparar el coeficiente de x,
2m = 1
metro = 1/2
metro + norte = 1
(1/2) + norte = 1
norte = 1/2
Ahora,
yo = ∫(1/2(2x + 1) + 1/2)/(x 2 + x + 3) dx
=1/2∫(2x + 1)/(x 2 + x + 1) dx + 1/2∫1/(x 2 + 2x(1/2) + (1/2)²-(1/2) ²+3) dx
= 1/2∫(2x + 1)/(x 2 + x + 1) dx + 1/2∫1/((x + 1/2) 2 + (11/4)) dx
= 1/2 ∫(2x + 1)/(x 2 + x + 1) dx + 1/2 ∫1/((x + 1/2) 2 + (√11/2) 2 ) dx
= 1/2 log|x 2 + x + 3| + (1/2) * (1/(√11/2)) bronceado -1 ((x + 1/2)/(√11/2)) + c
Como sabemos que ∫1/(x 2 + a 2 ) dx = 1/a tan -1 (x/a) + c
Por lo tanto, yo = 1/2 log|x 2 + x + 3| + 1/√11 bronceado -1 ((2x + 1)/(√11)) + c
Pregunta 3. ∫ (x – 3)/(x 2 + 2x – 4) dx
Solución:
Dado que I = ∫(x – 3)/(x 2 + 2x – 4) dx
Consideremos x – 3 = md/dx (x 2 + 2x – 4) + n
= m(2x + 2) + norte
x – 3 = (2m)x + (2m + n)
Al comparar los coeficientes de x,
2m = 1
metro = 1/2
2m + n = -3
2(1/2) + n = -3
n = -4
Asi que,
yo = ∫(1/2(2x + 2) – 4)/(x 2 + 2x – 4) dx
= 1/2 ∫(2x + 2)/(x 2 + 2x – 4) dx – 4∫ 1/(x 2 + 2x + (1) 2 – (1) 2 – 4) dx
= 1/2 ∫(2x + 2)/(x 2 + 2x – 4) dx – 4∫ 1/((x + 1) 2 – (√5)) dx
Como sabemos que ∫1/(x 2 – a 2 ) dx = 1/2a log|(x – a)/(x + a)| +c
= 1/2 log|x 2 + 2x – 4| – 4 × 1/(2√5) log|(x + 1 – √5)/(x + 1 + √5)| +c
Por lo tanto, yo = 1/2 log|x 2 + 2x – 4| – 2/√5 log|(x + 1 – √5)/(x + 1 + √5)| +c
Pregunta 4. ∫(2x – 3)/(x 2 + 6x + 13) dx
Solución :
Dado que I = ∫ (2x – 3)/(x 2 + 6x + 13) dx
Consideremos 2x – 3 = md/dx (x 2 + 6x + 13) + n
= m(2x + 6) + norte
2x – 3 = (2m)x + (6m + n)
Al comparar el coeficiente de x, obtenemos
2m = 2
metro = 1
6m + n = -3
6 * 1 + n = -3
n = -9
Ahora,
= ∫(1 * (2x + 6) – 9)/(x 2 + 6x + 13) dx
= ∫(2x + 6)/(x 2 + 6x + 13) dx + ∫(-9)/(x 2 + 2 * (3) * x + (3) 2 – (3) 2 + 13) dx
= ∫(2x + 6)/(x 2 + 6x + 13) dx -9 ∫1/((x + 3) 2 + (2)) dx
= logaritmo|(x 2 + 6x + 13)| – 9 * (1/2) bronceado -1 ((x + 3)/2) + c
Por lo tanto, I = log|(x 2 + 6x + 13)| – 9 × (1/2) bronceado -1 ((x + 3)/2) + c
Pregunta 5. ∫x 2 /(x 2 + 7x + 10) dx
Solución:
Dado que I = ∫x 2 /(x 2 + 7x + 10) dx
= ∫{1 – (7x + 10)/(x 2 + 7x + 10)}dx
yo = x – ∫(7x + 10)/(x 2 + 7x + 10) dx + c 1 ……..(i)
Sea I 1 = ∫(7x + 10)/(x 2 + 7x + 10) dx
Sea 7x + 10 = md/dx (x 2 + 7x + 10) + n
= m(2x + 7) + norte
7x + 10 = (2m)x + 7m + n
Al comparar los coeficientes de potencias iguales de x,
7 = 2m
metro = 7/2
7m + n = 10
7(7/2) + n = 10
n = -29/2
Entonces, l = ∫(1/6(6x – 4) – 1/3)/(3x 2 – 4x + 3) dx
= 1/6 ∫(6x – 4)/(3x 2 – 4x + 3) dx – 1/9 ∫1/(x 2 – 4/3x + 1) dx
= 1/6 ∫(6x – 4)/(3x 2 – 4x + 3) dx – 1/9 ∫1/(x 2 – 2x(2/3) + (2/3) 2 – (2/3) 2 + (2) 2 ) dx
= 1/6 ∫(6x – 4)/(3x 2 – 4x + 3) dx – 1/9 ∫1/(x – 2/3) 2 + (√5/2)) dx
= 1/6log|(3x 2 – 4x + 3) | – ((1/9) × 1/(√5/3)) tan((x – 2/3)/(√5/3)) + c
Por lo tanto, yo = 1/6log|(3x 2 – 4x + 3)| – (√5/15)tan -1 ((3x – 2)/√5) + c
Pregunta 6. ∫2x/(2 + x – x 2 ) dx
Solución:
Dado que I = ∫2x/(2 + x – x 2 ) dx
Ahora,
2x = m(d/dx(2 + x + x 2 )) + norte
2x = m(-2x + 1) + n
Ahora igualando el coeficiente de obtendremos m, n
m = -1,
norte = 1
∫2x/(2 + x – x 2 ) dx
= ∫(m(-2x + 1) + n)/(2 + x – x 2 ) dx
= ∫(-1(-2x + 1) + 1)/(2 + x – x 2 ) dx
= ∫(-1(-2x + 1))/(2 + x – x 2 ) dx + 1/(2 + x – x 2 ) dx
= -log|2 + x – x 2 | + ∫1/(2 + x – x 2 ) dx
= -log|2 + x – x 2 | – ∫1/(x 2 – x – 2) dx
= -log|2 + x – x 2 – ∫1/(x 2 – x(1/2)(2) + (1/2) – (1/2) – 2) dx
= -log|2 + x – x 2 | + ∫1/((x – 1/2) 2 – (3/2) 2 ) dx
= -log|2 + x – x 2 | – 1/3 log|((x – 1/2) – (3/2))/((x – 1/2) + (3/2))| +c
Por lo tanto, yo = -log|2 + x – x 2 | – 1/3 log|(x – 2)/(x + 1)| +c
Pregunta 7. ∫(1 – 3x)/(3x 2 + 4x + 2) dx
Solución:
Dado que I = ∫(1 – 3x)/(3x 2 + 4x + 2) dx
Consideremos 1 – 3x = md/dx (3x 2 + 4x + 2) + n
= m(6x + 4) + norte – 11 – 3x = (6m)× + (4λ + norte)
Al comparar los coeficientes de lx,
6m = -3
metro = -1/2
4m + norte = 1
4(-1/2) + norte = 1
norte = 3
yo = ∫(-1/2(6x + 4) + 3)/(3x 2 + 4x + 2) dx
= -1/2 ∫ (6x + 4)/(3x 2 + 4x + 2) dx + 3∫1/(3x 2 + 4x + 2) dx
= -1/2 ∫ (6x + 4)/(3x 2 + 4x + 2) dx + 3/3 ∫1/(x 2 + 4/3x + 2/3) dx
= -1/2 ∫ (6x + 4)/(3x 2 + 4x + 2) dx + ∫1/(x 2 + 2x(2/3) + (2/3) 2 – (2/3) 2 + (2/3) dia
= -1/2 ∫ (6x + 4)/(3x 2 + 4x + 2) dx + ∫1/((x + 2/3) 2 + 2/9) dx
= -1/2 ∫ (6x + 4)/(3x 2 + 4x + 2) dx + ∫1/((x + 2/3) 2 + (√2/3) 2 ) dx
= -1/2 log|(3x 2 + 4x + 2) | + 3/√2tan -1 ((x + 2/3)/(√2/3)) + c
Por lo tanto, yo = -1/2 log|(3x 2 + 4x + 2)| + 3/√2tan -1 ((3x + 2)/√2) + c
Pregunta 8. ∫(2x + 5)/(x 2 – x – 2) dx
Solución:
Dado que I = ∫(2x + 5)/(x 2 – x – 2) dx
Sea 2x + 5 = md/dx (x 2 – x – 2) + n
= m(2x – 1) + n
2x + 5 = (2m)x – m + n
Al comparar los coeficientes de x,
2m = 2
metro = 1
-m + n = 5
-1 + n = 5
norte = 6
Asi que,
yo = ∫((2x – 1) + 6)/(x 2 – x – 2) dx
= ∫ ((2x – 1))/(x 2 – x – 2) dx + 6∫1/(x 2 – 2x(1/2) + (1/2) 2 – (1/2) 2 – 2 ) dx
= ∫ (2x – 1)/(x 2 – x – 2) dx + 6∫1/((x – 1/2) 2 – 9/4) dx
= ∫ (2x – 1)/(x 2 – x – 2) dx + 6∫1/((x – 1/2) 2 – (3/2) 2 ) dx
= log|x 2 – x – 2| + 6/2(3/2) log|(x – 1/2 – 3/2)/(x – 1/2 + 3/2)| +c
Como sabemos que ∫1/(x 2 – a 2 ) dx = 1/2a log|(x – a)/(x + a)| +c
Por lo tanto, yo = log|x 2 – x – 2| + 2log|(x – 2)/(x + 1)| +c
Pregunta 9. ∫ (ax 3 + bx)/(x 4 + c 2 ) dx
Solución:
Dado que I = ∫(ax 3 + bx)/(x 4 + c 2 ) dx
Consideremos ax 3 + bx = md/dx (x 4 + c 2 ) + n
hacha 3 + bx = n(4x 3 ) + n
Al comparar los coeficientes de x,
4m = un
metro = un/4
y
norte = 0
yo = ∫(a/4 (4x 3 ) + bx)/(x 4 + c 2 ) dx
= a/4 ∫(4x 3 )/(x 4 + c 2 ) dx + b∫x/((x 2 ) 2 + c 2 ) dx
= a/4 ∫(4x 3 )/(x 4 + c 2 ) dx + b/2 ∫2x/((x 2 ) 2 + c 2 ) dx
= a/4 log|x 4 + c 2 | + b/2 I 1 ……..(i)
Ahora,
yo 1 = ∫2x/((x 2 ) 2 + c 2 ) dx
Ponga x 2 = t
2xdx = dt
yo 1 = ∫1/((t) 2 + c 2 ) dx
= 1/c tan -1 (t/c) + c 1
l 1 = 1/C tan -1 (x 2 /c) + c 1 …………(ii)
Ahora usando la ecuación (i) y (ii) obtenemos,
Por lo tanto, I = a/4 log|x 4 + c 2 | + b/2c tan(x 2 /c) + b
Pregunta 10. ∫(x + 2)/(2x 2 + 6x + 5) dx
Solución:
Dado que I = ∫(x + 2)/(2x 2 + 6x + 5) dx
Consideremos x + 2 = md/dx (2x 2 + 6x + 5) + n
= m(4x + 6) + n
x + 2 = (4m)x + (6m + n)
Al comparar los coeficientes de x,
Asi que,
4m = 1
metro = 1/4
6m + norte = 2
6(1/4) + norte = 2
norte = 1/2
yo = ∫(1/4(4x + 6)+1/2)/(2x 2 + 6x + 5) dx)
=1/4 ∫(4x + 6)/(2x 2 + 6x + 5) dx + 1/2 ∫1/(2x 2 + 6x + 5) dx
=1/4 ∫(4x + 6)/(2x 2 + 6x + 5) dx + 1/4 ∫1/(x 2 + 3x + 5/2) dx
=1/4 ∫(4x + 6)/(2x 2 + 6x + 5) dx + 1/4 ∫1/(x 2 + 2x(3/2) + (3/2) 2 – (3/2) 2 + 5/2) dx
=1/4 ∫(4x + 6)/(2x 2 + 6x + 5) dx + 1/4 ∫1/((x + 3/2) 2 + 1/4) dx
=1/4 ∫(4x + 6)/(2x 2 + 6x + 5) dx + 1/4 ∫1/((x + 3/2) 2 + (1/2) 2 ) dx
=1/4 ∫(4x + 6)/(2x 2 + 6x + 5) dx + 1/4 × (1/(1/2))tan -1 ((x + 3/2)/(1/ 2)) + c
Como sabemos que ∫1/(x 2 + a 2 ) dx = 1/a tan -1 (x/a) + c
Por lo tanto, yo = 1/4 log|2x 2 + 6x + 5| + 1/2 bronceado -1 (2x + 3) + c
Pregunta 11. ∫((3sinx – 2)cosx)/(5 – cos 2 x – 4sinx) dx
Solución:
Dado que I = ∫((3sinx – 2)cosx)/(5 – cos 2 x – 4sinx) dx
= ∫((3sinx – 2)cosx)/(5 – (1 – sin 2 x) – 4sinx) dx
= ∫((3senx – 2)senx)/(5 – 1 + sen 2 x – 4senx) dx)
Ahora sustituya sinx = t en la ecuación anterior
cosxdx = dt
Asi que,
yo = ∫(3t – 2)/(4 + t 2 – 4t) dt
= ∫((3t – 2))/(t 2 – 4t + 4) dt
= ∫(3t – 2)/(t – 2) 2 dt
Ahora integra fracciones parciales.
(3t – 2)/((t – 2) 2 ) = A/((t – 2)) + B/((t – 2) 2 )
= (A(t – 2) + B)/((t – 2) 2 )
= (En – 2A + B)/((t – 2) 2 )
3t – 2 = En – 2A + B
Al comparar los coeficientes tenemos, A = 3
y -2A + B = -2
Ahora, al sustituir el valor de A = 3 en la ecuación anterior,
-2 × 3 + B = -2
-6 + B = -2
B = 6 – 2
B = 4
Entonces, yo = ∫(3t – 2)/(t – 2) 2 dt
= ∫(3t – 2)/(t – 2) dt + ∫4/(t – 2) 2 dt
= 3log|t – 2| – 4(1/(t – 2)) + c
= 3log|t – 2| – 4(1/(t – 2)) + c
Ahora pon el valor de t = senx, tenemos ,
I = 3log|senx – 2| – 4(1/(senx – 2)) + c
Pregunta 12. ∫(5x – 2)/(1 + 2x + 3x 2 ) dx
Solución:
Dado que I = ∫(5x – 2)/(1 + 2x + 3x 2 ) dx
Consideremos 5x – 2 = A d/dx (1 + 2x + 3x 2 ) + B
5x – 2 = A(2 + 6x) + B
5x – 2 = 6 × A + 2A + B
Al comparar el Coeficiente tenemos, 6A = 5 y 2A + B = -2
A = 5/6
Al sustituir el valor de A en 2A + B = -2, tenemos n,
2 * 5/6 + B = -2
10/6 + B = -2
B = -2 – 10/6
B = (-12 – 10)/6
B = (-22)/6
B = (-11)/3
5x – 2 = 5/6(2 + 6x) – 11/3
Entonces, I = ∫(5x – 2)/(1 + 2x + 3x 2 ) dx se convierte en,
yo = ∫[5/6(2 + 6x) – 11/3]/(3x 2 + 2x + 1) dx
= 5/6∫(2 + 6x)/(3x 2 + 2x + 1) dx – 11/3∫dx/(3x 2 + 2x + 1)
= 5/6 log(3x 2 + 2x + 1) – 11/(3 × 3)∫dx/(x 2 + 2/3x + 1/3) + c
= 5/6 log(3x 2 + 2x + 1) – 11/9∫dx/(x 2 + 2/3 x + (4/3) 2 + 1/3 – (4/3) 2 ) + c
= 5/6 log(3x 2 + 2x + 1) – 11/9∫dx/((x + 1/3) 2 +(√2/3) 2 ) + c
= 5/6 log(3x 2 + 2x + 1) – 11/9 × 1/(√2/3) tan -1 (((x + 1/3)/(√2/3))] + C
Por lo tanto, I = 5/6 log(3x 2 + 2x + 1) – 11/(3√2) tan -1 [(3x + 1)/√2] + C
Publicación traducida automáticamente
Artículo escrito por anandchaturvedirishra y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA