Wyznacz całkę funkcji niewymiernej. Piotrek A.:

	x2
∫
	√2x2+2x+1

M:

Mariusz: Podstawienie Eulera

	x2		x2
∫		dx = √2∫		dx
	√2x2+2x+1		√4x2+4x+2

√4x²+4x+2 = t − 2x 4x²+4x+2 = t² − 4tx + 4x² 4x+2 = t² − 4tx 4tx+4x = t² − 2 x(4t+4) = t² − 2

	t2 − 2
x =
	4t+4

	4t2+4t − (2t2−4)
t − 2x =
	4t+4

	t2+2t+2
t − 2x = 2
	4t+4

	2t(4t+4) − 4(t2−2)
dx =		dt
	(4t+4)2

	8t2+8t−4t2+8
dx =		dt
	(4t+4)2

	t2+2t+2
dx = 4		dt
	(4t+4)2

	x2
∫		dx =
	√2x2+2x+1

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

	x2		(t2−2)2
∫		dx = 2√2∫		dt
	√2x2+2x+1		(4t+4)3

	x2		√2		(t2−2)2
∫		dx =		∫		dt
	√2x2+2x+1		32		(t+1)3

	x2		√2		t4−4t2+4
∫		dx =		∫		dt
	√2x2+2x+1		32		(t+1)3

	x2		√2		(t+1)4−4(t+1)3+2(t+1)2+4(t+1)+1
∫		dx =		∫		dt
	√2x2+2x+1		32		(t+1)3

	x2		√2
∫		dx =		(∫(t+1)dt −
	√2x2+2x+1		32

	dt		1		1
4∫dt+2∫		+4∫		dt+∫		dt)
	t+1		(t+1)2		(t+1)3

	x2
∫		dx =
	√2x2+2x+1

	√2		(t+1)2		4		1	1
		(		−4(t+1)+2ln(t+1)−		−			)
	32		2		t+1		2	(t+1)2

+C

	x2		√2		1	(t+1)4−1		(t+1)2+1
∫		dx =		(			−4		+2ln(t+1))
	√2x2+2x+1		32		2	(t+1)2		t+1

+C

	x2
∫		dx =
	√2x2+2x+1

	√2		(t2+2t+2)(t2+2t)		t2+2t+2
		(		−4		+2ln(t+1))
	32		2(t+1)*(t+1)		t+1

+C

	x2		√2		(t2+2t+2)(t2+2t)		t2+2t+2
∫		dx =		(		−2
	√2x2+2x+1		8		2(t+1)*4(t+1)		2t+2

	1
+		ln(t+1))+C
	2

	x2		√2		1
∫		dx =		((x+		)√4x2+4x+2−2√4x2+4x+2
	√2x2+2x+1		8		2

	1
+		ln(2x+1+√4x2+4x+2) )+C
	2

	x2		√2		1		1
∫		dx =		(		(2x−3)√4x2+4x+2 +		ln(2x+1+√4x2+4x+2) )
	√2x2+2x+1		8		2		2

+C

	x2		√2
∫		dx =		((2x−3)√4x2+4x+2 +ln(2x+1+√4x2+4x+2) )
	√2x2+2x+1		16

+C

	x2		1		√2
∫		dx =		(2x−3)√2x2+2x+1+		ln(2x+1+√2√2x2+2x+1) + C
	√2x2+2x+1		8		16