Add 1 to the coefficient and divide by that term. int[f(X) by the power rule is this.......X(n + 1)/n
int(6X)
= 6/2(X2)
= 3X2 + C
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A primitive to e^(x^(1/3)) is (e^(x^(1/3)))*(6-6x^(1/3)+3x^(2/3))
d/dx (ln(6x)) = (d/dx (6x))/(6x) = 6/6x = 1/x
8
6x-3y
2x -3 - 6x = 0 -3 = 6x - 2x -3 = 4x x = -3/4
integral 6dx over 4 to 10=[6x]410=6[10-4]=6*6=36
The integral of x^5 is (1/6)x^6 + C, where C is the constant of integration.
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
-6x
6x - 6x = 0
A primitive to e^(x^(1/3)) is (e^(x^(1/3)))*(6-6x^(1/3)+3x^(2/3))
The GCF is 6x.
-2
-6x^2 - 78x - 252 Improved answer: -6x+6x+7 When simplified = 7
1+6x+6x+8 1+ 12x +8 12x + 9 = 21 + x
6x - 9 = 21 6x = 21 + 9 6x = 28 x = 28/6 Reduce to: x = 14/3
There is no possible value of 'x' for which (6x-2) can be equal to (6x-12).