Jak policzyć taką całkę? wihajster: ^x2_√x²−x+1

jc:

	2t−t2
Podstawienie x=
	1−t2

prowadzi do całki

	2t2(2−t)2
−∫		dt
	(1−t2)3

czy jakoś podobnie

Mariusz: Mniej liczenia będzie jeśli zastosujemy pierwsze podstawienie √x²−x+1=t−x x²−x+1=t²−2tx+x² −x+1=t²−2tx 2tx−x=t²−1 x(2t−1)=t²−1

	t2−1
x=
	2t−1

	2t2−t−t2+1
t−x=
	2t−1

	t2−t+1
t−x=
	2t−1

	2t(2t−1)−2(t2−1)
dx=		dt
	(2t−1)2

	2t2−2t+2
dx=		dt
	(2t−1)2

	(t2−1)2	2t−1	2(t2−t+1)
∫				dt
	(2t−1)2	t2−t+1	(2t−1)2

	(t2−1)2
2∫		dt
	(2t−1)3

(16t⁴−32t²+16)−(2t−1)⁴ 16t⁴−32t²+16 16t⁴−32t³+24t²−8t+1 (32t³−56t²+8t+15)−4(2t−1)³ 32t³−56t²+8t+15 32t³−48t²+24t−4 (−8t²−16t+19)−(−2)(2t−1)² −8t²−16t+19 −8t²+8t − 2 −24t+21 −(−12)(2t−1) −24t+12 9 (2t−1)⁴+4(2t−1)³−2(2t−1)²−12(2t−1)+9

1		(2t−1)4+4(2t−1)3−2(2t−1)2−12(2t−1)+9
	∫		dt
8		(2t−1)3

1		dt		dt		dt
	(∫(2t−1)dt+4∫dt−2∫		−12∫		+9∫		)
8		2t−1		(2t−1)2		(2t−1)3

1		1		6		9	1
	(		(2t−1)2+2(2t−1)+		−			−ln|2t−1|)+C
8		4		2t−1		4	(2t−1)2