arctan x dx = Z

(1)

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Z

arctan x dx = Z

|{z} 1

=f ⁰

arctan x

| {z }

=g

dx = x

|{z}

=f

arctan x

| {z }

=g

− Z

|{z} x

=f

1 1 + x ²

| {z }

=g 0

dx

= x arctan x − 1 2

Z 2x

1 + x ² dx = x arctan x − 1

2 ln | 1 + x ²

| {z }

≥1>0

| + C

= x arctan x − 1

2 ln(1 + x ² ) + C, C ∈

^h

.

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g

d dx

x arctan x − 1

2 ln(1 + x ² ) + C

= (Dx) arctan x + xD arctan x

− 1 2

D(1 + x ² )

1 + x ² = arctan x + x 1 1 + x ² − 1

2 2x

1 + x ² = arctan x

ryab_Gr"_G|G|Ga

x ∈

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I =

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o

i-l/kmovu

2RJS M S = Dc

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s

V/rPXZ_e]

l(kmoGk

ala R b a f

l(kmEoek

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0 ≤ sin x ≤ 1

rNap_er2_G|e|a

x ∈ [0, π]

^k ^V(_e_GV ^i-l/k^nmovu

sin x − sin ² x

2 ≤ ln(1 + sin x) ≤ sin x

ryab_Gr2_e|G|a

x ∈ [0, π].

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id}pkG}mou

Z π 0

sin x − sin ² x 2

dx ≤

Z π 0

ln(1 + sin x) dx ≤ Z π

0

sin x dx.

_erPYa m(s

]P|e_]PV

i%l(kz{mou

π 0

− cos x = − cos π − (− cos 0) = − cos π + cos 0 = 1 + 1 = 2.

apc^Y0V m(s

]P|e_]PV

i%l(kmou

Z π 0

sin x dx − 1 2

Z π 0

sin ² x dx = 2 − 1 2

π 0

1 2 (x − sin x cos x) = 2 − π 4 .

_eX^Y0V

2 − π 4 ≤

Z π 0

ln(1 + sin x) dx ≤ 2.

2RJS

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\ZpV

y = ln x

^dac ^s ^]p\fabV

y = 1

^|eY_er2rya ^s ^c ^m ^_cXZYpg

ln x = 1 ⇔ x = e

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A = Z e

0

|1 − ln x| dx

| {z }

1,8 p.

= lim

a→0+

Z e a

|1 − ln x| dx

| {z }

0,5 p.

= lim

a→0+

Z e a

(1 − ln x) dx

| {z }

0,4 p.

.

(2)

Z e a

(1 − ln x) dx = Z e

a

1 dx − Z e

a

1 · ln x dx = e − a − ^e

a

x ln x + Z e

a

x · 1 x dx

= e − a − (e ln e − a ln a) + e − a = e − 2a + a ln a,

cd_G|G|G

ln e = 1

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a→0+ (e − 2a + a ln a) = e − 2 · 0 + 0 = e ≈ 2, 72.

−3

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−1 0 1 2

0 0.5 1 1.5 2 2.5 3

y

x

y=ln x y=1

s qya

}

g

\fbVV

f (x) = ln x

^danc

s

]p\Z_eYV

x = 0

^da

y = 1

\faydaPababV7ap|

s

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m

_eV"XfayWab|a

o

arctan x dx = Z

Z

arctan x dx = Z

|{z} 1

=f 0

arctan x

| {z }

=g

dx = x

|{z}

=f

arctan x

| {z }

=g

− Z

|{z} x

=f

1 1 + x 2

| {z }

=g 0

dx

= x arctan x − 1 2

Z 2x

1 + x 2 dx = x arctan x − 1

2 ln | 1 + x 2

| {z }

≥1>0

| + C

= x arctan x − 1

2 ln(1 + x 2 ) + C, C ∈

.

−

d dx

x arctan x − 1

2 ln(1 + x 2 ) + C

= (Dx) arctan x + xD arctan x

− 1 2

D(1 + x 2 )

1 + x 2 = arctan x + x 1 1 + x 2 − 1

2 2x

1 + x 2 = arctan x

x ∈

I =

ala R b a f

R b a f

f

0 ≤ sin x ≤ 1

x ∈ [0, π]

sin x − sin 2 x

2 ≤ ln(1 + sin x) ≤ sin x

x ∈ [0, π].

Z π 0

sin x − sin 2 x 2

dx ≤

Z π 0

ln(1 + sin x) dx ≤ Z π

0

sin x dx.

π 0

− cos x = − cos π − (− cos 0) = − cos π + cos 0 = 1 + 1 = 2.

Z π 0

sin x dx − 1 2

Z π 0

sin 2 x dx = 2 − 1 2

π 0

1

2 (x − sin x cos x) = 2 − π 4 .

2 − π 4 ≤

Z π 0

ln(1 + sin x) dx ≤ 2.

y = ln x

y = 1

ln x = 1 ⇔ x = e

A = Z e

0

|1 − ln x| dx

| {z }

1,8 p.

= lim

a→0+

=f ⁰

1 1 + x ²

1 + x ² dx = x arctan x − 1

2 ln | 1 + x ²

2 ln(1 + x ² ) + C, C ∈

2 ln(1 + x ² ) + C

D(1 + x ² )

1 + x ² = arctan x + x 1 1 + x ² − 1

1 + x ² = arctan x

sin x − sin ² x

sin x − sin ² x 2

sin ² x dx = 2 − 1 2

1 · ln x dx = e − a − ^e