Apparently that can't be solved with a finite number of so-called "elementary functions". You can get the beginning of the series expansion here: http://www.wolframalpha.com/input/?i=integrate+x^x
root x=x^(1/2) and 4x =4 x^1 you add the exponents then integrate as usual. The answer you should get is 4.
x/sqrt(x)=sqrt(x) integral sqrt(x)=2/3x3/2
Trying to integrate: cos2x sin x dx Substitute y = cos x Then dy = -sin x dx So the integral becomes: -y2dy Integrating gives -1/3 y3 Substituting back: -1/3 cos3x
By parts. You'll get ((x^2)/2) * (lnx - 1/2) + c
integrate(x5x dx) simplifies to integrate(5x^2 dx), and using the power rule of integration, add one to the power of x and divide the term by that number. Thus, x5x dx integrated is (5/3)x^3
x-1 = 1/x ∫1/x dx = ln x + C
If x has the power 2 then you want the integral of x2, I think. When you integrate this you get : x3/3 , plus a constant.
I can't integrate a-x /x-3 ?
{xe^x dx integrate by parts let f(x) = x so f'(x) = 1 and g(x) = e^x so g'(x) = e^x so.. f(x)*g(x) - {(g(x)*f'(x)) dx therefore xe^x - {(e^x * 1) dx so.. xe^x - e^x + C factorize so... (x-1)e^x + C
Apparently that can't be solved with a finite number of so-called "elementary functions". You can get the beginning of the series expansion here: http://www.wolframalpha.com/input/?i=integrate+x^x
-(10/x)
Integrate 2sin(x)cos(x)dxLet u = cos(x) and du = -sin(x)dx and pull out the -2:-2[Integral(u*du)]Integrate with respect to u:-2(u2)/2 + CSimplify:-u2 + CReplace u with cos(x):-cos2(x) + C
root x=x^(1/2) and 4x =4 x^1 you add the exponents then integrate as usual. The answer you should get is 4.
To integrate a function you find what the function you have is the derivative of. for example the derivative of x^2 is 2x. so the integral of 2x is x^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!
The integral of cot(x)dx is ln|sin(x)| + C