Can I get all the formulas of Indefinite Integration

Basic Formulas
∫ dx = x + C

The integral of a constant 1 with respect to x is simply x plus the constant of integration C.
∫ k dx = kx + C

When integrating a constant k, we multiply it by x and add C.
∫ xⁿ dx = (xⁿ⁺¹) / (n+1) + C, for n ≠ -1

This is the power rule for integration, where the exponent is increased by 1 and divided by the new exponent.
∫ (1/x) dx = ln|x| + C

This is a special case of the power rule when n = -1, leading to the natural logarithm function.
∫ e^x dx = e^x + C

The integral of the exponential function e^x is itself.
∫ a^x dx = (a^x / ln a) + C, for a > 0, a ≠ 1

The integral of an exponential function with base a follows this formula.
Trigonometric Integrals
∫ sin x dx = -cos x + C

The derivative of cos x is -sin x, so the integral of sin x is -cos x.
∫ cos x dx = sin x + C

The derivative of sin x is cos x, so its integral is sin x.
∫ sec² x dx = tan x + C

Since d/dx (tan x) = sec² x, its integral is tan x.
∫ cosec² x dx = -cot x + C

Since d/dx (cot x) = -cosec² x, the integral of cosec² x is -cot x.
∫ sec x tan x dx = sec x + C

Since d/dx (sec x) = sec x tan x, its integral is sec x.
∫ cosec x cot x dx = -cosec x + C

Since d/dx (cosec x) = -cosec x cot x, its integral is -cosec x.
Inverse Trigonometric Integrals
∫ 1 / (1 + x²) dx = tan⁻¹ x + C

This is derived from the derivative of tan⁻¹ x.
∫ 1 / √(1 - x²) dx = sin⁻¹ x + C

This follows from the derivative of sin⁻¹ x.
∫ -1 / √(1 - x²) dx = cos⁻¹ x + C

This follows from the derivative of cos⁻¹ x.
∫ 1 / (1 - x²) dx = 1/2 ln|(1 + x) / (1 - x)| + C

This integral is important in logarithmic transformations.
Special Functions and Properties
∫ log x dx = x log x - x + C

This is obtained using integration by parts.
∫ e^(ax) dx = (1/a) e^(ax) + C, for a ≠ 0

This follows from the chain rule.
∫ sinh x dx = cosh x + C

The derivative of cosh x is sinh x.
∫ cosh x dx = sinh x + C

The derivative of sinh x is cosh x.
Integration by Substitution
If we have an integral in the form of ∫ f(g(x)) g'(x) dx, we can substitute u = g(x), which simplifies the integral into ∫ f(u) du.

Integration by Parts
∫ u dv = uv - ∫ v du
This is useful when integrating a product of two functions.

These are the essential formulas for indefinite integration, providing a foundation for solving various integral problems. Let me know if you need further explanations!