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 + Cx^4:
x^4 → (x^5)/5 + C1/5 x^5 + Cx^2 and x^7:
x^2 → (x^3)/3 + Cx^7 → (x^8)/8 + Cx:
x → (x^2)/2 + Cx/4):
x/4 → 1/4 * (x^2)/2 + C → (x^2)/8 + C8x^3 → 2x^4 + C4, -7):
4 → 4x + C-7 → -7x + C*7x - 6:
7x - 6 → (7x^2)/2 - 6x + C6x^2 + 4x - 7:
6x^2 + 4x - 7 → 2x^3 + 2x^2 - 7x + C√x:
x^(1/2).(x^(3/2)) * (2/3) + Cx^(4/3) → (x^(7/3)) * (3/7) + Cx^(7/4) → (x^(11/4)) * (4/11) + C*d/dx [sin x] = cos x → ∫ cos x dx = sin x + Cd/dx [cos x] = -sin x → ∫ -sin x dx = cos x + Cd/dx [tan x] = sec^2 x→ ∫ sec^2 x dx = tan x + Cd/dx [cot x] = -csc^2 x → ∫ -csc^2 dx = cot x + Cd/dx [sec x] = sec x tan x → ∫ sec x tan x dx = sec x + Cd/dx [csc x] = -csc x cot x → ∫ csc x cot x dx = -csc x + C4 sin x - 5 cos x + 3 sec^2 x → -4 cos x - 5 sin x + 3 tan x + Cx, 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 + C4x 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)) + C1/x:
∫ (1/x) dx = ln|x| + C∫ (7/x) dx = 7 ln|x| + C∫ (1/(x + 5)) dx = ln|x + 5| + C