dx dx 3 sin x cos x dx dx x

1 dx ( )

∫

^x

x + 1

( )² ( )2 x

∫

^{x - 1} dx

3 sin x cos x dx

( )

∫

² ( )4 2x(x + 1 dx

∫

^{2 )4}

5 dx

( )

∫

^{2x - 1}

x - x + 5² ( )6

∫

^{log x}_x dx

1 t = x + 1 とおくと dt = dx ( )

 

∫

^x dx x + 1 ( )²

=

∫

^{t - 1} dt t²

=

∫

¹_t - dt¹ t²

= log|t| - + C = log|x + 1| -1 + C t

1 x + 1

2 t = とおくと

( ) x - 1   t = x - 1²   2t ⋅ dt = dx   x

∫

^{x - 1} dx

=

∫

(t + 1 t ⋅ 2t ⋅ dt^{2 )}

= 2 t + t dt

∫

^{4 2}

= t + t + C2 5 ⁵

2 3 ³

= 2 x - 1 + x - 1 + C

5( )² x - 1 2

3( ) x - 1 3 cos x = t とおくと、 - sin x dx = dt

( )

  sin x cos x dx

∫

²

=

∫

- t dt²

= - t + C = - cos x + C1 3 ³

1 3 ³

4 x + 1 = t とおくと、2x dx = dt ( ) ²

  2x(x + 1 dx

∫

^{2 )4}

=

∫

t dt⁴

= t + C = (x + 1 + C1 5 ⁵

1

5 ² ) 5 x - x + 5 = t とおくと、

( ) ²

  2x - 1 dx = dt( )  

∫

^{2x - 1} dx

x - x + 5²

=

∫

¹_t dt

= log|t| + C = log|x - x + 5| + C ²

6 log x = t とおくと、

( )

  dx = dt1 x

 

∫

^{log x}_x dx

=

∫

t dt

= t + C =1 log x + C 2 ²

1

2( )²

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1 x + 1 (x + 2x - 1 dx

( )

∫

^{( ) 2 )3 ( )2}

∫

^{2 x - 2( ) dx}

x - 4x² 3 tan x dx

( )

∫

( )4 (e + 1 (e + x dx

∫

^{x ) x )}

5 sin x dx

( )

∫

^{3 ( )6}

∫

_{x log x}^{1 dx}

1 x + 2x - 1 = t とおくと ( ) ²

  2x + 2 dx = dt( )   x + 1 dx = dt( ) 1

2

 

∫

⁽x + 1 (x + 2x - 1 dx^{) 2 )3}

=

∫

¹₂t dt³

= t + C = (x + 2x - 1 + C1 8 ⁴

1

8 ² )⁴

2 = t とおくと

( ) x - 4x²   x - 4x = t^{2 2}   2x - 4 dx = 2t dt( )  

∫

^{2 x - 2( )} dx

x - 4x²

=

∫

2 dt

= 2t + C = 2 x - 4x² + C 3 cos x = t とおくと、 - sin x dx = dt

( )

  tan x dx

∫

=

∫

^{sin x}_{cos x} dx

=

∫

¹_t dt

= -log|t| + C = -log|cos x| + C

4 e + x = t とおくと、(e + 1 dx = dt

( ) ^{x x} )

  (e + 1 (e + x dx

∫

^{x ) x )}

=

∫

t dt

= t + C = (e + x + C1 2 ²

1

2 ^x )

5 cos x = t とおくと、 - sin x dx = dt ( )

  sin x dx

∫

³

=

∫

sin x ⋅ sin x dx²

=

∫

sin x ⋅ (1 - cos x dx^{2 )}

=

∫

t - 1 dt²

= t - t + C = cos x - cos x + C1 3 ³

1 3 ³

6 log x = t とおくと、

( )

  dx = dt1 x

 

∫

_{x log x}¹ dx

=

∫

¹_t dt

= log|t| + C = log|log x| + C

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1 3 x + 1 x - 1 (x - 3x dx

( )

∫

^{( )( ) 3 )3} ( )2 sin x

∫

^{cos x} dx 3 dx

( )

∫

_{tan x}¹ ( )4

∫

^e dx

e + 1

x x

5 dx

( )

∫

^{log(log x}_x ⁾ ( )6 2 (2 + 1 dx

∫

^{x x )2}

1 x - 3x = t とおくと、(3x - 3 dx = dt

( ) ^{3 2} )

  3 x + 1 x - 1 (x - 3x dx

∫

^{( )( ) 3 )3}

=

∫

t dt³

= t + C = (x - 3x + C1 4 ⁴

1

4 ³ )⁴

2 = t とおくと、

( ) cos x   cos x = t²   - sin x = 2t ⋅ dt   sin x

∫

^{cos x} dx

=

∫

- 2t dt²

= - t + C = - cos x2 + C 3 ³

2

3 cos x

3 sin x = t とおくと、cos x dx = dt ( )

 

∫

_{tan x}¹ dx

=

∫

^{cos x}_{sin x} dx

=

∫

¹_t dx

= log|t| + C = log|sin x| + C

4 e + 1 = t とおくと、e dx = dt

( ) ^{x x}

 

∫

^e dx e + 1

x x

=

∫

¹_t dt

= log|t| + C = log(e + 1 + C^x )

5 log x = t とおくと、 dx = dt

( ) 1

x  

∫

^{log(log x}_x ⁾ dx

=

∫

log t dt

= t log t - t + C

= log x log log x - log x + C( )

6 2 + 1 = t とおくと、 dx = dt

( ) ^x 2

log 2

x

  2 (2 + 1 dx

∫

^{x x )2}

=

∫

_{log 2}^t2 dt

= t + C = + C

3 log 2

3 (2 + 1

3 log 2

x )³

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