Integrales Inmediatas ∫x
n
dx =
x ∫ a dx =
x n +1 +c n +1
ax +c ln a
∫ x dx = ln x + c
n=0
∫ dx = x + c
a=e
∫e
∫ Senxdx = −Cosx + c ∫ Cosxdx = Senx + c ∫ Tgxdx = − ln Cosx + c = ln Secx + c ∫ Ctgxdx = ln Senx + c ∫ Secxdx = ln( Secx + Tgx ) + c ∫ Cscxdx = ln( Cscx − Ctgx ) + c ∫ Sec xdx = Tgx + c ∫ Csc xdx = −Ctgx + c ∫ SecxTgxdx = Secx + c ∫ CscxCtgxdx = −Cscx + c
dx = ArcSenx + c 1− x2 1 ∫ 1 + x 2 dx = ArcTgx + c 1 ∫ x x 2 − 1 dx = ArcSecx + c
dx = e x + c
2
2
1
x
∫ Senhxdx = Coshx + c ∫ Coshxdx = Senhx + c ∫ Tghxdx = ln Coshx + c ∫ Ctghxdx = ln Senhx + c ∫ Sechxdx = ArcTgSenhx + c ∫ Cschxdx = − ArgThCoshx + c ∫ Sech xdx = Tghx + c ∫ Csch xdx = −Ctghx + c ∫ SechxTghxdx = − Sechx + c ∫ CschxCtghxdx = −Cschx + c
2
∫
1
n = −1
2
∫ ∫
1
( dx = ArgCoshx + c = ln ( x + −1
x +1 1 2
2
) − 1) + c
dx = ArgSenhx + c = ln x + x 2 + 1 + c x2
x 1 1 1+ x ∫ 1 − x 2 dx = ArgTghx + c = 2 ln 1 − x + c
1+ 1− x2 dx = − ArgSechx + c = − ln ∫ x 1− x2 x 1
+c