If you mean integral ln(x2+1)dx, it is 2tan-1(x)+x[ln(x2+1)-2]. To find this and other integrals, go to
http://integrals.wolfram.com/index.jsp?expr=ln(x^2%2B1)&random=false
It can solve nearly every integral I've encountered.
The integral of ln(x2) is 2x[ln(x) - 1] + C Do this using the method of integration by parts.
0.5
if you are integrating with respect to x, the indefinite integral of 1 is just x
The integral of 1 + x2 is x + 1/3 x3 + C.
The integral of a single term of a polynomial, in the form of AxN is (A/N+1) x (N+1). The first integral of 2x is x2 + C. The second integral of 2x is the first integral of x2 + C, which is 1/3x3 + Cx + C.
The integral of -x2 is -1/3 x3 .
In order to evaluate a definite integral first find the indefinite integral. Then subtract the integral evaluated at the bottom number (usually the left endpoint) from the integral evaluated at the top number (usually the right endpoint). For example, if I wanted the integral of x from 1 to 2 (written with 1 on the bottom and 2 on the top) I would first evaluate the integral: the integral of x is (x^2)/2 Then I would subtract the integral evaluated at 1 from the integral evaluated at 2: (2^2)/2-(1^2)/2 = 2-1/2 =3/2.
The indefinite integral of (1/x^2)*dx is -1/x+C.
The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
x=1
-1/15sin5ycos3y