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