Rewrite 7/5 int((x+a million)/(x^2+2x)). The by-made from the denominator is 2x+2. Multiply the numerator by skill of two and divide the vital by skill of two. we've 7/10 int((2x+2)/(x^2+2x)) it quite is interior the right variety to apply the ln variety. Ans: 7/10*ln(x^2+2x)
Rewrite 7/5 int((x+a million)/(x^2+2x)). The by-manufactured from the denominator is 2x+2. Multiply the numerator by using 2 and divide the crucial by using 2. we now have 7/10 int((2x+2)/(x^2+2x)) this is in the right variety to apply the ln variety. Ans: 7/10*ln(x^2+2x)
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Hola,
∫ x arcsenx dx =
sea:
x dx = dv → [1/(1+1)] x^(1+1) = (1/2)x² = v
arcsenx = u → [1 /√(1 - x²)] dx = du
luego, integrando por partes:
∫ u dv = v u - ∫ v du
∫ arcsenx x dx = (1/2)x² arcsenx - ∫ (1/2)x² [1 /√(1 - x²)] dx =
reescribamosla como:
(1/2)x² arcsenx + (1/2) ∫ [- x² /√(1 - x²)] dx (#)
cuanto a la restante integral, sumemos y sustraigamos 1 al numerador:
∫ [- x² /√(1 - x²)] dx =
∫ [(1 - x²) - 1] /√(1 - x²)} dx =
luego, partiendo y simplificando:
∫ [(1 - x²) /√(1 - x²)] dx - ∫ [1 /√(1 - x²)] dx =
∫ √(1 - x²) dx - ∫ [1 /√(1 - x²)] dx =
(siendo la segunda integral inmediata)
∫ √(1 - x²) dx - arcsenx =
ahora integremos la restante integral por partes, poniendo:
dx = dv → x = v
√(1 - x²) = u → (1 - x²)^(1/2) = u → (1/2)(- 2x)(1 - x²)^[(1/2) -1] dx =
- x (1 - x²)^(-1/2) dx = [- x /√(1 - x²)] dx = du
y obteniendo:
∫ u dv = v u - ∫ v du
∫ √(1 - x²) dx - arcsenx = {x √(1 - x²) - ∫ x [- x /√(1 - x²)] dx} - arcsenx =
x √(1 - x²) - ∫ [- x² /√(1 - x²)] dx - arcsenx
luego, resumiendo (ve sobre):
∫ [- x² /√(1 - x²)] dx = x √(1 - x²) - ∫ [- x² /√(1 - x²)] dx - arcsenx
teniendo la misma integral en ambos miembros, recojamosla al primer miembro:
∫ [- x² /√(1 - x²)] dx + ∫ [- x² /√(1 - x²)] dx = x √(1 - x²) - arcsenx + C
2 ∫ [- x² /√(1 - x²)] dx = x √(1 - x²) - arcsenx + C
por tanto:
∫ [- x² /√(1 - x²)] dx = (1/2) [x √(1 - x²) - arcsenx] + C = (1/2)x √(1 - x²) -
(1/2)arcsenx + C
entonces la expresión anterior (#) se vuelve:
(1/2)x² arcsenx + (1/2) ∫ [- x² /√(1 - x²)] dx = (1/2)x² arcsenx +
(1/2) [(1/2)x √(1 - x²) - (1/2)arcsenx] + C =
(1/2)x² arcsenx + (1/4)x √(1 - x²) - (1/4)arcsenx + C
en conclusión:
∫ x arcsenx dx = (1/2)x² arcsenx + (1/4)x √(1 - x²) - (1/4)arcsenx + C
espero haber sido de ayuda
¡Saludos!
Rewrite 7/5 int((x+a million)/(x^2+2x)). The by-made from the denominator is 2x+2. Multiply the numerator by skill of two and divide the vital by skill of two. we've 7/10 int((2x+2)/(x^2+2x)) it quite is interior the right variety to apply the ln variety. Ans: 7/10*ln(x^2+2x)
Rewrite 7/5 int((x+a million)/(x^2+2x)). The by-manufactured from the denominator is 2x+2. Multiply the numerator by using 2 and divide the crucial by using 2. we now have 7/10 int((2x+2)/(x^2+2x)) this is in the right variety to apply the ln variety. Ans: 7/10*ln(x^2+2x)
∫x*arcsin(x) dx
Use integration by parts:
u = arcsin(x)
du = dx/√1 - x²
dv = x dx
v = x²/2
uv - ∫v du
x²*arcsin(x)/2 - 1/2*∫x²/√1-x² dx
Substituting x = sin(t) in the integral.
x²*arcsin(x)/2 - 1/2*∫sin²(t) dt
x²*arcsin(x)/2 - 1/4[∫1 - cos(2t) dt]
x²*arcsin(x)/2 - t/4 + sin(t)*cos(t)/4 + C
x²*arcsin(x)/2 - arcsin(x)/4 + (x√1 - x²)/4 + C