# Integrales

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• Páginas : 10 (2261 palabras )
• Descarga(s) : 10
• Publicado : 24 de julio de 2010

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INTEGRALES
1)x5dx
x5dx = x⁶6+C
2)(x+x ) dx
(x+x ) dx =xdx +x1/2dx = x²2+23x3 +C
3)(3x -xx4 )dx
(3x -xx4 )dx =31x1/2dx-14x.x1/2dx
=3x-12dx -14x32dx=3 x-12+1-12+1 -14 x32+132+1+C
= 3x1/21/2 -14x5/25/2 +C =3.2x -14.25x5 +C
= 6x - 110x5 +C
4)x2x dx
x2x dx =x2x1/2 dx =x2x-1/2dx =x3/2dx = x32+132+1 + C = x5/25/2 +C = 2/5x5 +C
5) (1x2 + 4xx + 2) dx
(1x2 + 4xx + 2) dx =x-2dx + 41xx1/2dx + 2dx =x-2dx + 4x-3/2dx + 2dx=
x-2+1-2+1 + 4x-32+1-32+1 + 2x +C = -x-1+4x-1/2-1/2 + 2x +C = -1x - 81x + 2x +C
6)14x dx
14x dx = 1x1/4dx =x-1/4dx = x-14+1-14+1 + C =x3/43/4 +C =4/34x3 +C
7)(x2+13x )2 dx
(x2+13x )2 dx =(x2+1x1/3)2 dx = (x2+x-1/3)2 dx=(x4+2x5/3+x-2/3)dx=x4dx+2x5/3dx+x-2/3dx=
x55+2x53+153+1+x-23+1-23+1 +C = x55+2x8/38/3+x1/31/3+C =x55+343x8+33x+C

INTEGRACION POR SUSTITUCION
Recordar si u= f(x) → du= f´(x)dx
8)e5xdx
Cambio de variable 5x = u
5dx = du→ dx =15 du
e5xdx=eu15du=15eudu=15eu+C=15e5x+C
Generalizando: eax+bdx
Cambio de variable ax+b = ueax+bdx=eu1adu=1aeudu=1aeu+C=1aeax+b C
eax+bdx=1aeax+b+C
9)cos 5x dx
Recordar (sen x)´= cos x , cosx dx=sen x+C
C:V 5x =u
5dx =du →dx = 15du
cos5x dx=cosu 15du=15cosu du= 15sen u+C= 15sen 5x+C
Generalizando:cosax+bdx
c.v. ax+b =u cosax+bdx= cos u du=1acosu du=1asen u+C=sen ax+b+C
a dx =du→ dx = 1adu cos⁡(ax+b)dx=1asenax+b+C10)sen ax dx Recordar: (cos x)´= -sen x→sen x dx=-cosx+C
c.v. ax=u
sen ax dx =sen u1a du=1a sen u du =-1acosax+C
11)Lxxdx
c.v. Lx =u →1xdx=du
Lxxdx=Lx1xdx=u du=u22+C=(Lx)²2+C
12)1sen23xdx recordar: (cotg u)´=-1sen2uu´
c.v. cotg 3x = u-1sen23x3 dx=du
1sen23xdx=-13du
1sen23xdx=-13du=-13du=-13u+C=-13 cotg 3x+C
13)1cos27xdx Recordar (tag u)´= 1cos2uu´
c.v. tg 7x = u
1cos27x.7 dx=du → 1cos27xdx=17 du
1cos27xdx=17du=17du=17u+C=17tg 7x+C
14)13x-7dx
c.v. 3x-7=u
3 dx= du → dx=13 du
13x-7dx=1u.13du=131udu=13 ln|u|+C=13ln|3x-7|+C

15)11-xdx
c.v.1-x=u
-dx= du→dx== -du
11-xdx=1u (-1)du= -1u du=-lnu+C=-ln1-x+c
16)15-2xdx
c.v. 5-2x= u
-2dx = du→ dx=-12du
15-2xdx=1u-12du=-121udu=-12lnu+C=-12ln5-2x+C
Generalizando 14,15,16
1ax+bdx
1ax+bdx=1aln| ax+b|+C

17)tg 2x dx
tg2x= sen 2xcos2x
cv. cos 2x= u
-2sen 2x dx = du→ sen 2x dx = -12du
tg 2x dx=sen 2xcos2xdx=1u(-1/2 ) du= -121udu=-12ln|u|+C=-1 2ln|cos2x|+C

18) cotg5x-7dx
cotg (5x-7) = cos⁡(5x-7)sen(5x-7)
cv. sen(5x-7) = u
5cos (5x-7) dx= du → cos(5x-7) dx = 15du
cotg5x-7dx=cos⁡(5x-7)sen(5x-7)dx=1u15du=151udu=15ln|u|+C=15ln| sen 5x-7|+C

19)1cotg 3xdx
cotg 3x= cos3xsen 3x
cv.cos 3x = u
-3sen 3x dx= du → sen 3x dx = -1 3du
1cotg 3xdx=sen 3xcos3x dx=1u(-1 3 )du=-131udu=-13ln|u|+C=-13ln|cos3x|+C

20)sen2xcosx dx
cv. senx = u
cos x dx = du
u2 du= u33+C=sen3x3+C

21)cos3x sen x dx
cv. cos x = u
-sen x dx = du → sen x dx = -du
u3-1du=-u3du=-u44+C= -cos4x4+C

22)(x2+1).x dx
cv. x2+1=u
2x dx= du → x dx =1/2 du
u.12du=12u1/2du=12u12+112+1+C=12u3/23/2+C=12.23u3+C=13(x2+1)3+C

23)x2x2+3dx
cv.2x2+3=u
4x dx= du → x dx=14du
1u14du=14u-1/2du=14u-12+1-1/2+1+C=14u1/21/2+C=14.2u+C=122x2+3+C

24)x2x3+1dx
cv. x3+1=u
3x2dx=du→x2dx=13du
1u13du=13 u-1/2du=13u-12+1-12+1 +C=13u1/21/2+C=13.2u+C=23x3+1+C

25)cosxsen2xdx
cv. sen x=u
cos x dx = du
1u2du=u-2du=u-2+1-2+1+C=-u-1+C= -1 sen x+C...