Int (e^(-x^2)) = Int (1 + (-x^2) + (-x^2)^2/2! + (-x^2)^3/3! + ...
= x - x^3/3 + x^5/(5*2!) - x^7/(7*3!) ...
which, if taken with limits of integration from negative infinity to infinity, solves to the square root of x, making it one of the most famous and beautiful formulas in math.
maths signs
if you take your time youll figure out its e=mc2
tan(sqrtX) + C
x is negative 1x is negative 2
Use integration by parts. integral of xe^xdx =xe^x-integral of e^xdx. This is xe^x-e^x +C. Check by differentiating. We get x(e^x)+e^x(1)-e^x, which equals xe^x. That's it!
maths signs
-e^(-x) or negative e to the negative x this is because you multiply the function (e) by: 1 / (the derivative of the power ... in this case: -1) e^(-x) * (1/-1) = -e^(-x) Don't forget to add your constant!
-cotan(x)
integral of radical sinx
integral of e to the power -x is -e to the power -x
The indefinite integral of (1/x^2)*dx is -1/x+C.
arctan(x)
.5(x-sin(x)cos(x))+c
if you take your time youll figure out its e=mc2
the integral of the square-root of (x-1)2 = x2/2 - x + C
I'm not sure if you mean e^x + 17 or e^(x+17) so we'll do both. First, the integral of e^x + 17 because these terms are being added you can integrate them separately: integral((e^x)dx) + integral(17dx) integral of e^x is just e^x + C Integral of 17 is 17x + C, so we get: e^x + 17x + C Second, the integral of e^(x+17) we know how to integrate the form e^u, so just do a u substitution u=x+17 du=dx so we get integral((e^u)du)=e^u + C resubstitute for u and get e^(x+17) + C
x=1