limit x tends to infinitive ((e^x)-1)/(x)
replace square root o x with t.
tan(sqrtX) + C
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!
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(e^x)^8 can be written as e^(8*x), so the integral of e^(8*x) = (e^(8*x))/8 or e8x/ 8, then of course you have to add a constant, C.
limit x tends to infinitive ((e^x)-1)/(x)
integral (a^x) dx = (a^x) / ln(a)
The integral would be 10e(1/10)x+c
-cotan(x)
integral of radical sinx
integral of e to the power -x is -e to the power -x
better place to ask would be yahoo answers
replace square root o x with t.
A primitive to e^(x^(1/3)) is (e^(x^(1/3)))*(6-6x^(1/3)+3x^(2/3))
The indefinite integral of (1/x^2)*dx is -1/x+C.
This integral cannot be performed analytically. Ony when the integral is taken from 0 to infinity can it be computed by squaring the integral and applying a change of variable (switching to polar coordinates). if desired I could show how to do this.