x 2 + 4x + 5 dx,

(1)

"!#$%$'&(()*+$-,-/.01/$- " 2 2)3,%($'14.

57686:9 14;8(< 9=>5@? 5 AB =

,-1 6:6 <:&41 @CEDGFED=4H4HJI

(.0(<8.($-&((K): 2,*L+<8)J$%D

MN$/<:,%<E&O14$- ? ,-/.0&$%)P.(Q14,%,-R

AODTS:UWVYXWZT[]\"^YZ_W`aUWUWb][]^

c

Ued

Z 1

x ² + 4x + 5 dx,

c-f

d

Z

arctanx dx.

g8UhìXjUWVkîUlX2m(nlnUWViVYUWX"[G\oXWpeq/r(UeVYViUlbGUeVkX2m/VY[îm(bGpeVTrZ0ì[]sepW[GnUhbGbGUutkUnvWvhì[]îvlbwUWUjtx[G\

sjvhbG[zy{txpWbGbGUlVYZ|/vj^iZ0ZW}

= D c

Ued~Nb]Xep2ph^

a, b ∈

^y

a < b

^y:tkU

f : [a, b] →

àUjtxpe[îZ^kîm}2Z0bG[]îv(ynl[]Xjv3pW\

|4pe`Y`aUWVk'm(\(X"^i[]p4y

ala R ^b

a f

^tkU32bGv

R ^b

a f

}(o[Gb]bGpW[G\

f

pW\o[GZ0nUh\/\z[G\"^YZ_W`ipW[]^Ym(sjUe

c-f d~TVkpe[^aU(yZ0^k^iv

2 − π 4 ≤

Z ^π

0

ln(1 + sin x) dx ≤ 2.

FEDT@vevhì[^Y^YZb]ZtkUnvWvhì[]îvVkZ\âUWVYpeUWb]m(ZZ0\|([]\"âUjUhbwU(y{txpW\/X{UàUjtkUeU{s{Uh^Xjv{2àv

y = ln x

VYZ0XjvlVkm/pW`aUj^

x = 0

^tkU

y = 1

}([G[]ìàvVY[][]îvlX2m(sjU/}

D cUed~TVkpe[^aU(yZ0^k^iv

n→∞ lim Z ¹

0

x ⁿ 1 + x ⁶⁶⁶ sin √

x dx = 0.

c-f

dUWZX{UW[]X2X2[r[]OZ`YZ\"^Y[wUWUhbG[G"q"^avhbGW\

y ⁰ − t(1 + y ² ) = 0

`iUh^YXjUh[wVYm^sjvhbG[]bGbwv

[0, 1]

}8g:UW`YXjUWVkîUor(Z0ì[]sepW[GnlUWb]bwU(yEZ^kâv@bGj"îvhnveVk[X2m(sjUWm(XVkZ0^

pjsjUj^`iUh^YXjUh[wVYmjtkUs{vWb][GbGbGv

[0, 1]

^}

I

DTNUhZuXjUh[GX"X2[UWb]X2m/UW`Yseph^YZq"^iv{sjvh\

y ⁰⁰ − 5y ⁰ + 4y = sin t, y(0) = 0, y ⁰ (0) = 1,

`aUj^iX{UW[GVYm^sjvhbG[]bGbwv

}

oL(L()8):LER

(2)

}

D arctan x = _1+x ¹ 2

XjUh[GX2[]bGbGU

x ∈

^}

}

x ² + 4x + 5 = (x + 2) ² + 1

XjUh[GX2[]bGbwU

x ∈

^}

}

x − ^x 2 ² ≤ ln(1 + x) ≤ x

X{UW[]X2[Gb]bwU

x ≥ 0

^}

}epeV

f

pW\rZ`Y[GsWpe[^im(sjUN|/[GVk^YZZViVYv

g(x 0 )

^tkU

g

|([wVk^YZ0ZVYViv

x 0

yW\/[][G\

f ◦ g

pW\rZ`Y[GsWpe[^im(sjU

|([wVk^YZ0ZVYViv

x 0

tkU

(f ◦ g) ⁰ (x 0 ) = f ⁰ g(x 0 ) g ⁰ (x 0 )

^}

}

D ¹ ₂ (x − sin x cos x) = sin ² x

XjUW[]X2[Gb]bwU

x ∈

^}

}epeV

f

^tkU

g

pjsjUjûrZ0ì[]sepW[]^Ym(s2[wU|([wVxîZ0ZVYViv

x ∈

^yJ\([G[]\

f g

pe\>VY[][G\/vr(Z0`i[]sepW[]^Ym(sjUtkU

(f g) ⁰ (x) = Df (x)g(x) = f ⁰ (x)g(x) + f (x)g ⁰ (x)

^}

}Z

≈ 2, 72.

¡ }

lim x→0+ x ln x = 0.

¢ }

Dx ⁿ = nx ⁿ⁻¹

XjUh[GX2[]bGbGU

n ∈

^£ ^tkU

x ∈

^}

¤ }

D ln | x | = x ¹

X{UW[]X2[Gb]bwU

x ∈

\ { 0 }

^}

W }epeV

f : [a, b] →

pW\tkUj^iX"m/s{U/y(\([G[]\pe\peb]ZnlUeVYViU

ξ ∈ [a, b]

Vk[]^YZ\y(Z^Y^iv

Z ^b

a

f (x) dx = f (ξ)(b − a).

}epeV

a ≥ b > 0

^y\([G[G\

0 < a ¹ ≤ ¹ b

}

D sin x = cos x

^y

D cos x = − sin x

XjUh[GX2[]bGbwU

x ∈

^}

}~NbGXWp2ph^

p, q, r: ∆ →

tkUh^YX2m(s2[GU/}:epeV

y 1

tkU

y 2

pjs{Uh^XjUhXVY[bG[]\/ZUWUW`Y[wVYZVk^Y[`i[][G|(|(m(

nUjîpW\"îU+q(pWnlpe_WZ0Z\([wVkZ\¥"q"âvhbGW\

y ⁰⁰ + py ⁰ + qy = 0

`iUh^YXjUh[wVYm/UltkU

y e

pe\@txpeX"[G\

Z|/vhq(penlpW_WZZ0\([wVYZ0\2q"^ivWb]e\

y ⁰⁰ +py ⁰ + qy = r

`aUj^YXjUW[GVYmsjvWb][Gb]bwv

∆

yh\([G[G\X{UW[]X2X2[/Z|/vj

q(penlpW_WZZ0\([wVYZ0\@2q"îvWb]e\>àUjîX{UW[GVYmûsjvhbG[]bGbGv

∆

^{pjsjUh^}

y = y e + C 1 y 1 + C 2 y 2 ,

^nl[wVYViv

C 1 , C 2 ∈

^}

}

0 ≤ sin x ≤ 1

XjUh[GX"[GbGbGU

x ∈ [0, π]

^}

}

ln

^Z

= 1

^}

}

arctan:

→ ] − ^π 2 , ^π ₂ [

^pW\ ^f ^[txZX"^Y[Gp/}

¡ }¦um(sjUWm(XVkZ0^

sin, cos:

→

pjs{Uh^bG[G\(ZUeUhì[GVYZVk^Y[ì[][G|(|(m/nlUh^kîpWnl[GU/}

¢ }

3 · 5 = 15

^}

h¤ }

25 + 9 = 34

^}

x 2 + 4x + 5 dx,

Z 1

x 2 + 4x + 5 dx,

Z

arctanx dx.

a, b ∈

a < b

f : [a, b] →

ala R b

a f

R b

a f

f

2 − π 4 ≤

Z π

0

ln(1 + sin x) dx ≤ 2.

y = ln x

x = 0

y = 1

n→∞ lim Z 1

0

x n 1 + x 666 sin √

x dx = 0.

y 0 − t(1 + y 2 ) = 0

[0, 1]

[0, 1]

y 00 − 5y 0 + 4y = sin t, y(0) = 0, y 0 (0) = 1,

D arctan x = 1+x 1 2

x ∈

x 2 + 4x + 5 = (x + 2) 2 + 1

x ∈

x − x 2 2 ≤ ln(1 + x) ≤ x

x ≥ 0

f

g(x 0 )

g

x 0

f ◦ g

x 0

(f ◦ g) 0 (x 0 ) = f 0 g(x 0 ) g 0 (x 0 )

D 1 2 (x − sin x cos x) = sin 2 x

x ∈

f

g

x ∈

f g

(f g) 0 (x) = Df (x)g(x) = f 0 (x)g(x) + f (x)g 0 (x)

≈ 2, 72.

lim x→0+ x ln x = 0.

Dx n = nx n−1

n ∈

x ∈

D ln | x | = x 1

x ∈

\ { 0 }

f : [a, b] →

ξ ∈ [a, b]

Z b

a

f (x) dx = f (ξ)(b − a).

a ≥ b > 0

0 < a 1 ≤ 1 b

D sin x = cos x

D cos x = − sin x

x ∈

p, q, r: ∆ →

y 1

y 2

y 00 + py 0 + qy = 0

y e

y 00 +py 0 + qy = r

∆

∆

y = y e + C 1 y 1 + C 2 y 2 ,

C 1 , C 2 ∈

0 ≤ sin x ≤ 1

x ∈ [0, π]

ln

= 1

x ² + 4x + 5 dx,

ala R ^b

R ^b

Z ^π

n→∞ lim Z ¹

x ⁿ 1 + x ⁶⁶⁶ sin √

y ⁰ − t(1 + y ² ) = 0

y ⁰⁰ − 5y ⁰ + 4y = sin t, y(0) = 0, y ⁰ (0) = 1,

D arctan x = _1+x ¹ 2

x ² + 4x + 5 = (x + 2) ² + 1

x − ^x 2 ² ≤ ln(1 + x) ≤ x

(f ◦ g) ⁰ (x 0 ) = f ⁰ g(x 0 ) g ⁰ (x 0 )

D ¹ ₂ (x − sin x cos x) = sin ² x

(f g) ⁰ (x) = Df (x)g(x) = f ⁰ (x)g(x) + f (x)g ⁰ (x)

Dx ⁿ = nx ⁿ⁻¹

D ln | x | = x ¹

Z ^b

0 < a ¹ ≤ ¹ b

y ⁰⁰ + py ⁰ + qy = 0

y ⁰⁰ +py ⁰ + qy = r

→ ] − ^π 2 , ^π ₂ [