x2

5

7

x2+0,5

x2

∫(6

x5 − 4

)dx = 6∫x

2 dx − 4∫x2−0,5dx = 6

− 4

+C =

x

7

2,5

12

8 x2

12 x3 x − 8 x2

2

=

x7 −

x +C =

x +C.

7

5

7

5

Пример 2.

1−3x

1 −3x =t, 3x =1 −t

1

t

1

t

1

1−3x

x =1 −t , dx = −1 dt

∫e

dx =

= −

3 ∫e

dt

= −

3 e

+C

= −3 e

+C.

3

3

Пример 3.

6x −1 = t, 6x = t +1

∫

dx

=

= 1

∫

dt

=

1 tgt +C =

1 tg(6x −1) +C.

t +1

dt

cos

2

1)

x =

,

dx =

2

t

(6x −

6

6

6

cos

6

6

Пример 4.

ln(x + 7) = t,

∫ln(x + 7) dx =

dx

= ∫tdt = t2

+C =

1 ln2 (x + 7) +C.

′

x + 7

dt = (ln(x +

7)) dx =

2

2

x

+ 7

1.

∫

ln xdx =

u = ln x,

du =

dx

= xln x −

∫

x dx = xln x

−

∫

dx = xln x − x +C =

x

dv = dx,

v = x

x

= x(ln x −1) +C

2.∫(2x +1) cos3xdx =

u =

2x +1,

du = 2dx

2x +1

2

1 sin 3x

dv = cos3xdx,

v =

=

3

sin 3x − 3 ∫sin 3xdx =

3

=

2x +1sin 3x +

2 cos3x +C.

3

9

3.∫x2 sin xdx =

u = x2 ,

du = 2xdx

= −x2 cos x + 2∫xcos xdx =

dv = sin xdx,

v = −cos x

=

u = x,

du = dx

= −x2 cos x + 2(xsin x − ∫sin xdx) =

dv = cos xdx,

v = sin x

= −x2 cos x + 2xsin x + 2cos x +C.

Аудиторные задания.

Найти неопределенные интегралы

1.∫(1 −3x) ln(4x)dx,

2.∫(2x +3) cos5xdx,

3.∫(1 − x2 )sin xdx,

4.∫(5x2 +1)e−xdx.

Вариант 1.

1.

∫(3 − x) sin 4xdx.

2.

∫(x2 − 4) cos xdx.

3.

∫x ln(2x)dx.

Вариант 2.

1.

∫(1 − x)сos2xdx.

2.

∫(3 −x2 ) sin xdx.

3.

∫(x +3) ln xdx.

Вариант 3.

1.

∫(x − 2) sin 3xdx.

2.

∫x2 cos xdx.

3.

∫ln(1 −3x)dx.

Вариант 4.

1.

∫(x2 + 2x +3)ex dx

2.

∫(x + 6) cos xdx.

3.

∫ln(x +1)dx.