zadanka imie lub nick: Witam, Potrzebuję pomocy w kilku zadaniach. Jak obliczyć pochodną funkcji: y=lnlnlnx całki: ∫cos(4x²+8)dx

	x5+2x3+4x+4
∫		dx
	x4+2x3+2x2

	cos2xdx
∫
	cos2x

	xdx
∫
	(x+1)12+(x+1)13

	3x3dx
∫
	√x2+4x+5

∫(e^−x+1)³dx ∫cos²xdx granice:

	sin 2n
lim→∞ p3{
	1n

lim→∞arctg(^sinx_x)

Adamm:

	1		1		1
(ln(ln(lnx)))'=		*(ln(lnx))'=		*		*(lnx)'=
	ln(lnx)		ln(lnx)		lnx

	1		1		1
=		*		*
	ln(lnx)		lnx		x

imie lub nick: oo super! A dalej jak robić? Możesz wytlumaczyć?

Adamm:

	1		3
∫(e−x+1)3dx=∫e−3x+3e−2x+3e−x+1 dx = −		e−3x−		e−2x−3e−x+x+c
	3		2

Adamm:

	1		x		sin(2x)
∫cos2xdx=		∫1+cos(2x)dx =		+		+c
	2		2		4

Adamm: lim ³√sin(2/n)/(1/n) = lim ³√2*sin(2/n)/(2/n) = ³√2

Adamm:

	sinx
limx→∞ arctg(		) = arctg(0) = 0
	x

piotr:

x5 + 2 x3 + 4 x + 4
	=
x4 + 2 x3 + 2 x2

	4 + 4 x + 4 x2 + 4 x3
=−2 + x +
	x4 + 2 x3 + 2 x2

piotr: (x⁵ + 2 x³ + 4 x + 4)/(x⁴ + 2 x³ + 2 x²) = = −2 + 2/x² + x + (2 (1 + 2 x))/(2 + 2 x + x²)

piotr: ∫(−2 + 2/x² + x + (2 (1 + 2 x))/(2 + 2 x + x²)) dx = = −(2/x) − 2 x + x²/2 − 2 arctg[1 + x] + 2 ln[2 + 2 x + x²]

Adamm:

	3x3		1
∫		dx=(ax2+bx+c)√x2+4x+5+A∫		dx
	√x2+4x+5		√x2+4x+5

3x3		x+2		A
	=(2ax+b)√x2+4x+5+(ax2+bx+c)		+
√x2+4x+5		√x2+4x+5		√x2+4x+5

3x³=(2ax+b)(x²+4x+5)+(ax²+bx+c)(x+2)+A 3a−3=0, 10a+2b=0, 10a+6b+c=0, A+5b+2c=0 a=1, b=−5, c=20, A=15

	3x3		1
∫		dx=(x2−5x+20)√x2+4x+5−15∫		dx
	√x2+4x+5		√x2+4x+5

	1		1
∫		dx=∫		dx
	√x2+4x+5		√(x+2)2+1

x+2=sinht dx=coshtdt

	cosht
∫		dt=∫dt = t = asinh(x+2)+c
	√sinh2t+1

	3x3
∫		dx=(x2−5x+20)√x2+4x+5−15asinh(x+2)+c
	√x2+4x+5

imie lub nick:

	−2+2x2+x+(2(1+2x))
∫		dx = U{−(2x)−2x+x2}{2−2arctg[1 + x]
	(2+2x+x2))

+2ln[2+2x+x²]

imie lub nick:

	−(2x)−2x+x2
=
	2−2arctg[1 + x]+2ln[2+2x+x2]

imie lub nick: tak to powinno wyglądać?

Adamm:

	x
∫		dx
	√x+1+3√x+1

x+1=t⁶ dx=6t⁵dt

	t6−1		t9−t3
6∫		*t5dt = 6∫		dt
	t3+t2		t+1

	t9+t8−t8−t7+t7+t6−t6−t5+t5+t4−t4−t3
=6∫		dt = 6∫t8−t7+t6−t5+t4−t3dt =
	t+1

	2		3		6		6		3
=		t9−		t8+		t7−t6+		t5−		t4+c =
	3		4		7		5		2

	2		3		6		6
=		6√x+19−		6√x+18+		6√x+17−6√x+16+		6√x+15−
	3		4		7		5

	3
−		6√x+14+c
	2

imie lub nick: Wow, wielkie dzięki!

Mariusz:

	x5+2x3+4x+4		x5+2x3+4x+4
∫		dx=∫		dx
	x4+2x3+2x2		x2(x2+2x+2)

	x5+2x3+4x+4		x5+x3		4x+4
∫		dx=∫		dx+∫		dx
	x2(x2+2x+2)		x2(x2+2x+2)		x2(x2+2x+2)

	x3+x		(2x2+4x+4)−2x2
=∫		dx+∫		dx
	x2+2x+2		x2(x2+2x+2)

	x3+x−2		2
=∫		+∫		dx
	x2+2x+2		x2

	3x+2		2
=∫(x−2)dx+		dx+∫		dx
	(x+1)2+1		x2

x−2 x³+x−2:x²+2x+2 x³+2x²+2x −2x²−x−2 −2x²−4x−4 3x+2

	3x3
∫		dx
	√x2+4x+5

√x²+4x+5=t−x x²+4x+5=t²−2tx+x² 4x+5=t²−2tx 2tx+4x=t²−5 x(2t+4)=t²−5

	t2−5
x=
	2t+4

	2t2+4t−t2+5		t2+4t+5
t−x=		=
	2t+4		2t+4

	2t(2t+4)−2(t2−5)
dx=		dt
	(2t+4)2

	2t2+8t+10
dx=		dt
	(2t+4)2

	(t2−5)3	2t+4	2(t2+4t+5)
3∫				dt
	(2t+4)3	t2+4t+5	(2t+4)2

	(t2−5)3
6∫		dt
	(2t+4)4

3		(t2−5)3
	∫		dt
8		(t+2)4

∫cos²xdx=sin(x)cos(x)+∫sin²(x)dx ∫cos²xdx=sin(x)cos(x)+∫(1−cos²(x))dx ∫cos²xdx=sin(x)cos(x)+∫dx−∫cos²xdx 2∫cos²xdx=sin(x)cos(x)+x+C₁

	1
∫cos2xdx=		(sin(x)cos(x)+x)+C
	2