X^2? 1/3x^3
Int = 3x^(2) dy y = 3x^(3) / 3 + c y = x^(3) + C
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
x/3 - 1/x = 2/3find a common denominator: 3x(x*x)/3x - (1*3)/3x = 2/3(x2-3)/(3x) = 2/3(x2-3)*(3) = (3x)*(2)3x2-9 = 6x3x2-6x-9 = 0x2-2x-3 = 0(x-3)(x+1)=0x-3=0 & x+1=0x=3 & x=-1
-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