Evalúa las siguientes integrales:
Pregunta 1. ∫(x + 1)√(x 2 – x + 1)dx
Solución:
Tenemos,
∫(x + 1)√(x 2 – x + 1)dx
Sea x + 1 = ad(x 2 – x + 1)/dx + b
=> x + 1 = a(2x – 1) + b
Al comparar ambos lados, obtenemos,
=> 2a = 1 yb – a = 1
=> a = 1/2 y b = 1 + 1/2 = 3/2
Entonces, nuestra ecuación se convierte en,
= ∫[(1/2)(2x – 1) + 3/2]√(x 2 – x + 1)dx
= (1/2)∫(2x – 1)√(x 2 – x + 1)dx + (3/2)∫√(x 2 – x + 1)dx
Para la primera parte, sea x 2 – x + 1 = t, entonces tenemos, (2x – 1)dx = dt
Entonces tenemos,
= (1/2)∫√tdt + (3/2)∫√[(x – 1/2) 2 + (√3/2) 2 ]dx
= (1/2)(2/3)(t 3/2 ) + (3/2)∫√[(x – 1/2) 2 + (√3/2) 2 ]dx
= (1/3)(x 2 – x + 1) 3/2 + (3/2)[(1/2)(x – 1/2)√(x 2 – x + 1) + (3/8 )log|(x – 1/2) + √(x 2 – x + 1)|] + c
= (1/3)(x 2 – x + 1) 3/2 + (3/8)(2x – 1)√(x 2 – x + 1) + (9/16) log|(x – 1/ 2) + √(x 2 – x + 1)| +c
Pregunta 2. ∫(x + 1)√(2x 2 + 3)dx
Solución:
Tenemos,
∫(x + 1)√(2x 2 + 3)dx
Sea x+1 = ad(2x 2 + 3)/dx + b
=> x + 1 = a(4x) + b
Al comparar ambos lados, obtenemos,
=> 4a= 1 y b = 1
=> a = 1/4 y b = 1
Entonces, la ecuación se convierte en,
= ∫[(1/4)(4x) + 1]√(2x 2 + 3)dx
= ∫(1/4)(4x)(x + 1)√(2x 2 + 3)dx + ∫√(2x 2 + 3)dx
Para la primera parte, sea 2x 2 + 3 = t, entonces tenemos, 4xdx = dt
Entonces tenemos,
= (1/4)∫√tdt + √2∫√(x 2 +3/2)dx
= (1/4)(2/3)(t 3/2 ) + √2[(x/2)√(x 2 + 3/2) + (3/4) log |x + √(x 2 + 3/2)|] + c
= (1/6)(2x 2 + 3) 3/2 + (x/2)√(2x 2 + 3) + (3/2√2) log |x + √(x 2 + 3/2)| +c
Pregunta 3. ∫(2x – 5)√(2 + 3x – x 2 )dx
Solución:
Tenemos,
∫(2x – 5)√(2 + 3x – x 2 )dx
Sea 2x – 5 = ad(2 + 3x – x 2 )/dx + b
=> 2x – 5 = a(3 – 2x) + b
Al comparar ambos lados, obtenemos,
=> –2a= 2 yb + 3a = –5
=> a = –1 yb = –5 – 3(–1) = –2
Entonces, la ecuación se convierte en,
= ∫[(–1)(3 – 2x) – 2]√(2 + 3x – x 2 )dx
= –∫(3 – 2x)√(2 + 3x – x 2 )dx – 2∫√(2 + 3x – x 2 )dx
Para la primera parte, sea 2 + 3x – x 2 = t, entonces tenemos, (3 – 2x)dx = dt
Entonces tenemos,
= –∫√tdt – 2∫√[(17/4) – (9/4 – 3x – x 2 )]dx
= –(2/3)(t 3/2 ) – 2∫√[(√17/2) 2 – (x – 3/2) 2 ]dx
= –(2/3)(2 + 3x – x 2 ) 3/2 – 2[(1/2)(x – 3/2)√(2 + 3x – x 2 ) + (17/8) sen – 1 [(x – 3/2)/(√17/2)]] + c
= –(2/3)(2 + 3x – x 2 ) 3/2 – (1/2)(2x – 3)√(2 + 3x – x 2 ) – (17/8) sen -1 [(2x – 3)/√17] + do
Pregunta 4. ∫(x + 2)√(x 2 + x + 1)dx
Solución:
Tenemos,
∫(x + 2)√(x 2 + x + 1)dx
Sea x + 2 = ad(x 2 + x + 1)/dx + b
=> x + 2 = a(2x + 1)+b
Al comparar ambos lados, obtenemos,
=> 2a = 1 y a + b = 2
=> a = 1/2 y b = 2 – 1/2 = 3/2
Entonces, la ecuación se convierte en,
= ∫[(1/2)(2x + 1) + 3/2]√(x 2 + x + 1)dx
= (1/2)∫(2x + 1)√(x 2 + x + 1)dx + (3/2)∫√(x 2 + x + 1)dx
Para la primera parte, sea x 2 + x + 1 = t, entonces tenemos, (2x + 1)dx = dt
Entonces tenemos,
= (1/2)∫√tdt + (3/2)∫√[(x + 1/2) 2 + (√3/2) 2 ]dx
= (1/2)(2/3)(t 3/2 ) + (3/2)[(1/2)(x + 1/2)√(x 2 + x + 1) + (3/8 ) log|(x + 1/2) + √(x 2 + x + 1)|] + c
= (1/3)(x 2 + x + 1) 3/2 + (3/8)(2x + 1)√(x 2 + x + 1) + (9/16) log|(x + 1/ 2) + √(x 2 + x + 1)| +c
Pregunta 5. ∫(4x + 1)√(x 2 – x – 2)dx
Solución:
Tenemos,
∫(4x + 1)√(x 2 – x – 2)dx
Sea 4x + 1 = ad(x 2 – x – 2)/dx + b
=> 4x + 1 = un (2x – 1) + segundo
Comparando ambos lados, obtenemos,
=> 2a = 4 y b – a = 1
=> a = 2 y b = 1 + 2
=> a = 2 y b = 3
Entonces, la ecuación se convierte en,
= ∫[2(2x – 1) + 3]√(x 2 – x – 2)dx
= 2∫(2x – 1)√(x 2 – x – 2)dx + 3∫√(x 2 – x – 2)dx
Para la primera parte, sea x 2 – x – 2 = t, entonces tenemos, (2x – 1)dx = dt
Entonces tenemos,
= 2∫√tdt + 3∫√(x 2 – x – 2)dx
= 2∫√tdt + 3∫√[(x – 1/2) 2 – (3/2) 2 ]dx
= 2(2/3)(t 3/2 ) + 3[(1/2)(x – 1/2)√(x 2 – x – 2) – (9/8) log|(x – 1/ 2) + √(x 2 – x – 2)|] + c
= (4/3)(x 2 – x – 2) 3/2 + (3/4)(2x – 1)√(x 2 – x – 2) – (27/8) log|(x – 1/ 2) + √(x 2 – x – 2)| +c
Pregunta 6. ∫(x – 2)√(2x 2 – 6x + 5)dx
Solución:
Tenemos,
∫(x – 2)√(2x 2 – 6x + 5)dx
Sea x – 2 = ad(2x 2 – 6x + 5)/dx + b
=> x – 2 = a(4x – 6) + b
Al comparar ambos lados, obtenemos,
=> 4a = 1 y b – 6a = –2
=> a = 1/4 y b = –2 + 6(1/4)
=> a = 1/4 y b = –1/2
Entonces, la ecuación se convierte en,
= ∫[(1/4)(4x – 6) + (–1/2)]√(2x 2 – 6x + 5)dx
= (1/4)∫(4x – 6)√(2x 2 – 6x + 5)dx – (1/2)∫√(2x 2 – 6x + 5)dx
Para la primera parte, sea 2x 2 – 6x + 5 = t, entonces tenemos, (4x – 6)dx = dt
Entonces tenemos,
= (1/4)∫√tdt – (√2/2)∫√(x 2 – 3x + 5/2)dx
= (1/4)(2/3)(t 3/2 ) – (√2/2)∫√[(x – 3/2) 2 + (1/2) 2 ]dx
= (1/6)(2x 2 – 6x + 5) 3/2 – (1/√2)[(1/2)(x – 3/2)√(x 2 – 3x + 5/2) + ( 1/8) log|(x – 3/2) + √(x 2 – 3x + 5/2)|] + c
= (1/6)(2x 2 – 6x + 5) 3/2 – (1/8)(2x – 3)√(2x 2 – 6x + 5) – (1/8√2) log|(x – 3/2) + √(x2 – 3x + 5/2)| +c
Pregunta 7. ∫(x + 1)√(x 2 + x + 1)dx
Solución:
Tenemos,
∫(x + 1)√(x 2 + x + 1)dx
Sea x + 1 = ad(x 2 + x + 1)/dx + b
=> x + 1 = a(2x + 1)+b
Al comparar ambos lados, obtenemos,
=> 2a = 1 y a + b = 1
=> a = 1/2 y b = 1/2
Entonces, la ecuación se convierte en,
= ∫[(1/2)(2x + 1) + 1/2]√(x 2 + x + 1)dx
= (1/2)∫(2x + 1)√(x 2 + x + 1)dx + (1/2)∫√(x 2 + x + 1)dx
Para la primera parte, sea x 2 + x + 1 = t, entonces tenemos, (2x + 1)dx = dt
Entonces tenemos,
= (1/2)∫√tdt + (1/2)∫√[(x + 1/2) 2 + (√3/2) 2 ]dx
= (1/2)(2/3)(t 3/2 ) + (1/2)[(1/2)(x + 1/2)√(x 2 + x + 1) + (3/8 ) log|(x + 1/2) + √(x 2 + x + 1)|] + c
= (1/3)(x 2 + x + 1) 3/2 + (1/8)(2x + 1)√(x 2 + x + 1) + (3/16)log|(x + 1/ 2) + √(x 2 + x + 1)| +c
Pregunta 8. ∫(2x + 3)√(x 2 + 4x + 3)dx
Solución:
Tenemos,
∫(2x + 3)√(x2 + 4x + 3)dx
Sea 2x + 3 = ad(x 2 + 4x + 3)/dx + b
=> 2x + 3 = a(2x + 4) + b
Al comparar ambos lados, obtenemos,
=> 2a = 2 y 4a + b = 3
=> a = 1 y b = 3 – 4 = –1
Entonces, la ecuación se convierte en,
= ∫[2x + 4 + (–1)]√(x 2 + 4x + 3)dx
= ∫(2x + 4)√(x 2 + 4x + 3)dx – ∫√(x 2 + 4x + 3)dx
Para la primera parte, sea x 2 + 4x + 3 = t, entonces tenemos, (2x + 4)dx = dt
Entonces tenemos,
= ∫√tdt – ∫√(x 2 + 4x + 3)dx
= (2/3)(t 3/2 ) – ∫√[(x + 2) 2 –1]dx
= (2/3)(x 2 + 4x + 3) 3/2 – (1/2)(x + 2)√(x 2 + 4x + 3) + (1/2)log |x + 2 + √ (x2 + 4x + 3)| +c
Pregunta 9. ∫(2x – 5)√(x 2 – 4x + 3)dx
Solución:
Tenemos,
∫(2x – 5)√(x 2 – 4x + 3)dx
Sea 2x – 5 = ad(x 2 – 4x + 3)/dx + b
=> 2x – 5 = a(2x – 4) + b
Al comparar ambos lados, obtenemos,
=> 2a = 2 yb – 4a = –5
=> a = 1 y b = –5 + 4(1) = –1
Entonces, la ecuación se convierte en,
= ∫2x – 4 + (–1)]√(x 2 – 4x + 3)dx
= ∫(2x + 4)√(x 2 – 4x + 3)dx – ∫√(x 2 – 4x + 3)dx
Para la primera parte, sea x 2 – 4x + 3 = t, entonces tenemos, (2x – 4)dx = dt
Entonces tenemos,
= ∫√tdt – ∫√(x 2 – 4x + 3)dx
= (2/3)(t 3/2 )– ∫√[(x – 2) 2 – 1]dx
= (2/3)(x 2 – 4x + 3) 3/2 – (1/2)(x – 2)√(x 2 – 4x + 3) + (1/2)log |(x – 2) + √(x 2 – 4x + 3)| +c
Pregunta 10. ∫x√(x 2 + x)dx
Solución:
Tenemos,
∫x√(x 2 + x)dx
Sea x = ad(x 2 + x)/dx + b
=> x = a(2x + 1) + b
Al comparar ambos lados, obtenemos,
=> 2a = 1 y a + b = 0
=> a = 1/2 y b = –1/2
Entonces, la ecuación se convierte en,
= ∫[(1/2)(2x + 1) + (–1/2)]√(x 2 + x)dx
= (1/2)∫(2x + 1)√(x 2 + x)dx – 1/2∫√(x 2 + x)dx
Para la primera parte, sea x 2 + x = t, entonces tenemos, (2x + 1)dx = dt
Entonces tenemos,
= (1/2)∫√tdt – 1/2∫√(x 2 + x)dx
= (1/2)(2/3)(t 3/2 )– (1/2)∫√[(x + 1/2) 2 – (1/2) 2 ]dx
= (1/3)(t 3/2 ) – 1/2[(1/2)(x + 1/2)√(x 2 + x)] + (1/8)log|(x + 1/ 2) + √(x 2 + x)|] + c
= (1/3)(x 2 + x) 3/2 – (1/8)(2x + 1)√(x 2 + x) + (1/16)log|(x + 1/2) + √ (x2 + x)| +c
Pregunta 11. ∫(x – 3)√(x 2 + 3x – 18)dx
Solución:
Tenemos,
∫(x – 3)√(x2 + 3x – 18)dx
Sea x – 3 = ad(x 2 + 3x – 18)/dx + b
=> x – 3 = a(2x + 3) + b
Al comparar ambos lados, obtenemos,
=> 2a = 1 y 3a + b = –3
=> a = 1/2 y b = –3 – 3/2 = –9/2
Entonces, la ecuación se convierte en,
= ∫[(1/2)(2x + 3) – (9/2)]√(x 2 + 3x – 18)dx
= (1/2)∫(2x + 3)√(x 2 + 3x – 18)dx – (9/2)∫√(x 2 + 3x – 18)dx
Para la primera parte, sea x 2 + 3x – 18 = t, entonces tenemos, (2x + 3)dx = dt
Entonces tenemos,
= (1/2)∫√tdt – (9/2)∫√(x 2 + 3x – 18)dx
= (1/2)(2/3)(t 3/2 ) – (9/2)∫√[(x + 3/2) 2 – (9/2) 2 ])dx
= (1/3)(x 2 + 3x – 18) 3/2 – (9/2)[(x + 3/2)√(x 2 + 3x – 18) – (81/8) log |(x + 3/2) + √(x 2 + 3x – 18)|] + c
= (1/3)(x 2 + 3x – 18) 3/2 – (9/8)(2x + 3)√(x 2 + 3x – 18) + (729/16) log |(x + 3/ 2) + √(x 2 + 3x – 18)| +c
Pregunta 12. ∫(x + 3)√(3 – 4x – x 2 )dx
Solución:
Tenemos,
∫(x + 3)√(3 – 4x – x 2 )dx
Sea x + 3 = ad(3 – 4x – x 2 )/dx + b
=> x + 3 = a(–4 – 2x) + b
Al comparar ambos lados, obtenemos,
=> –2a = 1 yb – 4a = 3
=> a = –1/2 y b = 3 + 4(–1/2) = 1
Entonces, la ecuación se convierte en,
= ∫[(–1/2)(–4 – 2x) + 1]√(3 – 4x – x 2 )dx
= (–1/2)∫(–4 – 2x)√(3 – 4x – x 2 )dx + ∫√(3 – 4x – x 2 )dx
Para la primera parte, sea 3 – 4x – x 2 = t, entonces tenemos, (–4 – 2x)dx = dt
Entonces tenemos,
= (–1/2)∫√tdt + ∫√(3 – 4x – x 2 )dx
= (–1/2)(2/3)(t 3/2 ) + ∫√[(√7) 2 – (x + 2) 2 ]dx
= (–1/3)(3 – 4x – x 2 ) 3/2 + (1/2)[(x + 2)√(3 – 4x – x 2 ) + 7 tan –1 [(x + 2) /√7]] + c
= (–1/3)(3 – 4x – x 2 ) 3/2 + (1/2)(x + 2)√(3 – 4x – x 2 ) + (7/2) tan –1 [(x + 2)/√7] + c
Publicación traducida automáticamente
Artículo escrito por gurjotloveparmar y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA