Int = 3x^(2) dy
y = 3x^(3) / 3 + c
y = x^(3) + C
square root x
(1/8)(x-sin 4x)
replace square root o x with t.
No. A square number is one that has an integral as its square root. For example, 9 is a square number because 3 x 3 = 9. There is no whole number that gives you 11 when multiplied by itself.
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
The general form is (2/3)*x^(3/2) + C
the integral of the square-root of (x-1)2 = x2/2 - x + C
square root x
sqrt(x^3) (x^3)^.5 x^1.5 1.5+1=2.5 1/2.5=.4 Answer: .4x^2.5+c
(1/8)(x-sin 4x)
replace square root o x with t.
No. A square number is one that has an integral as its square root. For example, 9 is a square number because 3 x 3 = 9. There is no whole number that gives you 11 when multiplied by itself.
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
The integral of -x2 is -1/3 x3 .
(ex)3=e3x, so int[(ex)3dx]=int[e3xdx]=e3x/3 the integral ex^3 involves a complex function useful only to integrations such as this known as the exponential integral, or En(x). The integral is:-(1/3)x*E2/3(-x3). To solve this integral, and for more information on the exponential integral, go to http://integrals.wolfram.com/index.jsp?expr=e^(x^3)&random=false
S sqrt(X - 1)/x dx this, I think, can be rewritten as.............. S (X - 1)^1/2 * 1/X dx 2/3(X -1)^3/2 * ln(X) + C 2/3ln(X)/sqrt[(X -1)^3] + C
There is no formula relating to a perfect square but if you want a method 1. Find square root(x) 2. Take the integer component (integral value) of square root(x) 3 Add 1 to intenger(square root(x)) 4. square it So: (integer(square root(x)) + 1)^2