To integrate e^(-2x)dx, you need to take a u substitution.
u=-2x
du=-2dx
Since the original integral does not have a -2 in it, you need to divide to get the dx alone.
-(1/2)du=dx
Since the integral of e^x is still e^x, you get:
y = -(1/2)e^(-2x)
Well, that was one method. I usually solve easier functions like this by thinking how the function looked like before it was differentiated. I let f(x) stand for the given function and F(x) stand for the primitive function; the function we had before differentiation (the integrated function).
f(x)= e-2x <-- our given function
F(x)= e-2x/-2 <-- our integrated function
Evidence: F'(x)= -2e-2x/-2 = e-2x = f(x) Q.E.D
It's as simple as that.
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e^x/1-e^x
use the double angle formula for cos(2x) which is: cos(2x)=2cos^2(x)-1 by this relation cos^2(x)=(cos(2x)+1)/2 now we'd integrate this instead this will give sin(2x)/4+x/2 =) hope this helps
you need to know natural logarithms3e to the 2x-1 power = 8(2x-1) ln e = ln (8/3)ln e = 1(2x-1) = ln(8/3) = 0.982x = 1.98x = 0.99
f(x) = (x^2)(e^x)f'(x) = e^x((x^2)+2x) - i thinkf"(x) = ?--------f(x) = (x^2)(e^x)apply the power rulef'(x) = (x^2)(e^x) + (2x)(e^x)apply the power rule to the first part and apply the power rule to the second part, then add those togetherf''(x) = [(x^2)(e^x) + (2x)(e^x)] + [(2x)(e^x) + (2)(e^x)]simplifyf''(x) = (e^x)(x^2 + 4x +2)I got it right. It checked out on my calculator.
int(ln(x2)dx)=xln|x2|-2x int(ln2(x)dx)=x[(ln|x|-2)ln|x|+2]