The 3s would cancel and it would become the integral of 1/x which is ln x.
X^2? 1/3x^3
Int = 3x^(2) dy y = 3x^(3) / 3 + c y = x^(3) + C
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.
multiply the number times x. For example, the integral of 3 is 3x.
If you mean integral[(2x^2 +4x -3)(x+2)], then multiply them out to get: Integral[2x^3+8x^2+5x-6]. This is then easy to solve and is = 2/4x^4+8/3x^3+5/2x^2-6x +c
-3x
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
X^2? 1/3x^3
dy/dx = 3 integral = (3x^2)/2
(3x-1) * exp(3*x) / 9
Int = 3x^(2) dy y = 3x^(3) / 3 + c y = x^(3) + C
5
multiply the number times x. For example, the integral of 3 is 3x.
-cos(3x) + constant
7
If you mean integral[(2x^2 +4x -3)(x+2)], then multiply them out to get: Integral[2x^3+8x^2+5x-6]. This is then easy to solve and is = 2/4x^4+8/3x^3+5/2x^2-6x +c