0
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.
What else can I help you with?
Integral of -3 dx?
-3x
What is the integral of 3sin3x?
-cos(3x) + constant
Integral of 3 raised to x?
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.
What is the integral of -3dx?
∫(-3)dx = -3x + C
How do you Differentiate and Integrate y equals 3x?
dy/dx = 3 integral = (3x^2)/2
Can 3x-4 be simplified?
7
What is the integral of expx2?
X^2? 1/3x^3
What is integral of xe3x?
(3x-1) * exp(3*x) / 9
What is the integral of 3 over 3x?
The 3s would cancel and it would become the integral of 1/x which is ln x.
Integral of 3x-2 dx?
5
Determine the integral 3x minus1 x plus 2dx?
Int[3x-x+2] =Int[2x+2] =x^2 +2x +C
What is the indefinite answer of 3(3x-1)4 dx?
To find the indefinite integral of the expression (3(3x-1)^4 , dx), we can use the power rule of integration. First, we can factor out the constant 3, then apply the substitution method or directly integrate. The integral becomes: [ \int 3(3x-1)^4 , dx = 3 \cdot \frac{(3x-1)^5}{15} + C = \frac{(3x-1)^5}{5} + C, ] where (C) is the constant of integration.
