całka amcavoy: Jak obliczyć ∫₀¹ [x^3m + x^2m + x^m]*[2x^2m + 3x^m + 6]^1/m dx

wakacje: pomysł na początek zaczerpnięty z internetu: (x^3m+x^2m+x^m)(2x^2m+3x^m+6)^1/m= =x(x^3m−1+x^2m−1+x^m−1)(2x^2m+3x^m+6)^1/m= =(x^3m−1+x^2m−1+x^m−1)*x(2x^2m+3x^m+6)^1/m= =(x^3m−1+x^2m−1+x^m−1)*[x^m(2x^2m+3x^m+6)]^1/m= =(x^3m−1+x^2m−1+x^m−1)*(2x^3m+3x^2m+6x^m))^1/m t=2x^3m+3x^2m+6x^m → dt=6m(x^3m−1+x^2m−1+x^m−1)dx ∫(x^3m+x^2m+x^m)(2x^2m+3x^m+6)^1/mdx= =∫((x^3m−1+x^2m−1+x^m−1)*(2x^3m+3x^2m+6x^m))^1/m)dx=

	1
=		∫t1/mdt=
	6m

	1		t(1/m)+1
=		*		=
	6m		1   +1 m

	1		t(1/m)+1
=		*		=
	6m		m+1   m

	1		m
=		*t(1/m)+1		=
	6m		m+1

	t(1/m)+1		(2x3m+3x2m+6xm)(1/m)+1
=		=
	6(m+1)		6(m+1)

stąd: ₀¹∫((x^3m+x^2m+x^m)(2x^2m+3x^m+6)^1/m)dx=

	(2x3m+3x2m+6xm)(1/m)+1
=[		]01=
	6(m+1)

	(2*13m+3*12m+6*1m)(1/m)+1
=(		)−
	6(m+1)

	(2*03m+3*02m+6*0m)(1/m)+1
−(		)=
	6(m+1)

	11(1/m)+1
=		−0=
	6(m+1)

	11(1/m)+1
=
	6(m+1)