Integraly Zakladne Vzorce

  • April 2020
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Z

Z

dx =

Z

xa dx =

Z

dx = ln |x| + c, x

Z

f 0 (x) dx = ln |f (x)| + c, f (x)

Z

eax e dx = + c, a

Z

ax a dx = + c, ln a

Z

cos ax sin ax dx = − + c, a

Z

sin ax + c, cos ax dx = a

Z

dx tg ax = + c, 2 cos ax a

Z

cotg ax dx + c, =− 2 a sin ax

1 dx = x + c,

for x ∈ R.

xa+1 + c, a+1

for a ∈ R, a 6= −1, x ∈ R − {0}. for x ∈ R − {0}.

ax

x

for f (x) 6= 0, x ∈ D(f ). Z

for a ∈ R, a 6= 0, x ∈ R, Z

for a > 0, a 6= 1, x ∈ R,

ex dx =

for a ∈ R, a 6= 0, x ∈ R,

Z

for a ∈ R, a 6= 0, x ∈ R − {kπ ; k ∈ Z},

cos x dx = sin x + c. Z

o π for a ∈ R, a 6= 0, x ∈ R − (2k + 1) ; k ∈ Z , 2 n

Z

ex + c = ex + c. ln e

sin x dx = − cos x + c. Z

for a ∈ R, a 6= 0, x ∈ R,

ex dx = ex + c.

dx = tg x + c. cos2 x

dx = − cotg x + c. sin2 x

dx x x for a ∈ R, a 6= 0, x ∈ (− |a| ; |a|). + c1 = − arccos + c2 , = arcsin 2 |a| |a| −x Z √ dx √ = ln x + x2 − a2 + c, for a ∈ R, a 6= 0, x ∈ (−∞ ; − |a|) ∪ (|a| ; ∞). 2 2 x −a Z   √ dx √ = ln x + x2 + a2 + c, for a ∈ R, a 6= 0, x ∈ R. x2 + a2 Z x 1 x dx 1 = arctg + c1 = − arccotg + c2 , for a ∈ R, a 6= 0, x ∈ R. 2 2 x +a a a a a   Z Z x − a 1 1 dx 1 1 + c, = − ln dx = for a ∈ R, a 6= 0, x ∈ R − {±a}. x2 − a2 2a x − a x + a 2a x + a Z Z cosh ax + c, sinh ax dx = for a ∈ R, a 6= 0, x ∈ R, sinh x dx = cosh x + c. a Z Z sinh ax cosh ax dx = for a ∈ R, a 6= 0, x ∈ R, cosh x dx = sinh x + c. + c, a Z Z dx tgh ax dx + c, = for a ∈ R, a 6= 0, x ∈ R, = tgh x + c. 2 a cosh ax cosh2 x Z Z cotgh ax dx dx + c, =− = − cotgh x + c. for a ∈ R, a 6= 0, x ∈ R − {0}, 2 a sinh ax sinh2 x Z



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