Jul 19, 2024
f(x) = x^n
, then f'(x) = nx^(n-1)
.F'(x) = x^n
, then F(x) = (x^(n+1))/(n+1) + C
.C
.3x^2
:
3x^2
→ x^3 + C
x^4
:
x^4
→ (x^5)/5 + C
1/5 x^5 + C
x^2
and x^7
:
x^2
→ (x^3)/3 + C
x^7
→ (x^8)/8 + C
x
:
x
→ (x^2)/2 + C
x/4
):
x/4
→ 1/4 * (x^2)/2 + C
→ (x^2)/8 + C
8x^3
→ 2x^4 + C
4
, -7
):
4 → 4x + C
-7 → -7x + C
7x - 6
:
7x - 6
→ (7x^2)/2 - 6x + C
6x^2 + 4x - 7
:
6x^2 + 4x - 7
→ 2x^3 + 2x^2 - 7x + C
√x
:
x^(1/2)
.(x^(3/2)) * (2/3) + C
x^(4/3)
→ (x^(7/3)) * (3/7) + C
x^(7/4)
→ (x^(11/4)) * (4/11) + C
d/dx [sin x] = cos x
→ ∫ cos x dx = sin x + C
d/dx [cos x] = -sin x
→ ∫ -sin x dx = cos x + C
d/dx [tan x] = sec^2 x
→ ∫ sec^2 x dx = tan x + C
d/dx [cot x] = -csc^2 x
→ ∫ -csc^2 dx = cot x + C
d/dx [sec x] = sec x tan x
→ ∫ sec x tan x dx = sec x + C
d/dx [csc x] = -csc x cot x
→ ∫ csc x cot x dx = -csc x + C
4 sin x - 5 cos x + 3 sec^2 x
→ -4 cos x - 5 sin x + 3 tan x + C
x
, plus C
.∫ 6x^2 dx
→ 2x^3 + C
∫ from 1 to 2 of 6x^2 dx
→ 14
∫ from a to b f(x) dx = F(b) - F(a)
d/dx [e^u] = e^u * u'
∫ e^u du = e^u / u' + C
∫ e^x dx = e^x + C
∫ e^5x dx = (e^5x)/5 + C
4x e^(x^2) dx → 2 e^(x^2) + C
∫ (1/x^2) dx
→ -1/x + C
∫ (1/x^3) dx
→ - (1/2x^2) + C
∫ (8/x^4) dx
→ - (8/3x^3) + C
∫ (1/(4x - 3)^2) dx → - (1/(4(4x - 3))) + C
∫ (7/(5x - 3)^4) dx → - (7/(15(5x - 3)^3)) + C
1/x
:
∫ (1/x) dx = ln|x| + C
∫ (7/x) dx = 7 ln|x| + C
∫ (1/(x + 5)) dx = ln|x + 5| + C