In order to compute that integral, we need to use the power rule:
∫ xⁿ dx = xn + 1/(n + 1) + c where n is any constant except 0 and -1.
Apply that rule to get:
∫ 3x dx
= 3 ∫ x dx [Factor out the constant]
= 3 ∫ x1 dx [Make note of the exponent]
= 3x1 + 1/(1 + 1) + c
= 3x2/2 + c
So that is the integral of 3x.
-3x
-cos(3x) + constant
The integral of 3x is ln(3)*3x. Take the natural log of the base and multiply it by the base raised to the power.
∫(-3)dx = -3x + C
dy/dx = 3 integral = (3x^2)/2
7
X^2? 1/3x^3
(3x-1) * exp(3*x) / 9
The 3s would cancel and it would become the integral of 1/x which is ln x.
5
Int[3x-x+2] =Int[2x+2] =x^2 +2x +C
multiply the number times x. For example, the integral of 3 is 3x.