całka wojtek:

	1
oblicz całkę nieoznaczoną z ∫		dx
	√1−x2+√x2+1

jc:

	√x2 +1 − √x2 − 1
= ∫		dx = ...
	x2

jc: x = sh t

	√x2+1		ch2 t
∫		dx = ∫		dt = ∫ [1 − (1 − cth2 t)] dt = t − cth t
	x2		sh2 t

Podobnie liczymy drugą całkę.

Mariusz:

	√1−x2−√x2+1
∫		dx
	(1−x2)−(x2+1)

	√1−x2		√x2+1
−∫		dx+∫		dx
	2x2		x2

	√1−x2
−∫		dx
	2x2

√(1−x²)=(x+1)t (1−x)(1+x)=(x+1)²t² 1−x=(x+1)t² 1−x=xt²+t² 1−t²=x+xt² x(1+t²)=1−t²

	1−t2		−1−t2+2		2
x=		=		=−1+
	1+t2		1+t2		1+t2

	2t
(x+1)t=
	1+t2

dx=2(−1)(1+t²)⁻²2tdt

	4t
dx=−		dt
	(1+t2)2

	4t	(1+t2)2	t
∫				dt
	(1+t2)	(1−t2)2	(1+t2)2

	4t2
∫		dt
	(1+t2)(1−t2)2

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

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

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

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

	1+t2
∫		dt−arctan(t)
	(1−t2)2

	(1−t)2+2t)
∫		dt−arctan(t)
	(1−t2)2

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

	1		1
=−		+		−arctan(t)
	1+t		1−t2

	1	√1−x2		√1−x2
=			−arctan(		)+C
	2	x		x+1

	√x2+1
∫		dx
	x2

√x²+1=t−x x²+1=t²−2tx+x² 1=t²−2tx 2tx=t²−1

	t2−1
x=
	2t

	2t2−t2+1		t2+1
t−x=		=
	2t		2t

	2t*2t−2(t2−1)
dx=		dt
	4t2

	t2+1
dx=		dt
	2t2

	4t2	(t2+1)	(t2+1)
∫				dt
	(t2−1)2	2t	2t2

	(t2+1)2
∫		dt
	t(t2−1)2

	(t2−1)2+4t2
∫		dt
	t(t2−1)2

	dt		4t
∫		+∫		dt
	t		(t2−1)2

	2
=ln|t|−		+C
	t2−1

	√x2+1
=−		+ln|x+√x2+1|+C
	x

	√1−x2		√x2+1
−∫		dx+∫		dx
	2x2		x2

1	√1−x2		√x2+1		√1−x2
		−		−arctan(		)+ln|x+√x2+1|+C
2	x		x		x+1