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angel.white · #10 May 11th, 2008, 06:49 PM

Quote:

Originally Posted by Hibijibi

And second one is xe^(-x^2) dx

Practice just looking at these ones and being able to tell the answer. First thing you should see is that x is the correct power of a derivative of x^2. And second you should see that e^(whatever) will differentiate to e^(whatever) so

e^(-x^2) differentiates to e^(-x^2) and then we apply the chain rule and get -2xe^(-x^2)

And when you can see this easily, all you need to do is realize that if you multiply this by -1/2 then you will have your integral.

ie: (-1/2) *(-2xe^(-x^2)) = xe^(-x^2)

and since -1/2 is a constant, it will carry all the way through from a f(x) to f'(x)

so we simply change e^(-x^2) into (-1/2)e^(-x^2)

and now when you differentiate it, you will get the formula you are trying to integrate.

meaning $\frac d{dx} \left(-\frac 12 ~e^{-x^2}\right) = x~e^{-x^2}$

integrate both sides to get:
$-\frac 12 ~e^{-x^2} = \int x~e^{-x^2} dx$

Here are some more, try doing them without substitution (look at it to find the answer):

1. $\int 2 ~cos(x) ~e^{sin(x)}~dx$

2. $\int \frac {arctan(x)}{x^2+1}~dx$

hint for #2: $\frac d{dx} arctan(x) = \frac 1{x^2+1}$

3. $\int \frac {arctan^2(x)}{x^2+1}~dx$

4. $\int \frac {1}{(x^2+1)~arctan(x)}~dx$

hint for #4: $\frac d{dx} ln(x) = \frac 1x$ this hint is a little abstract, if you can't figure out how to apply the thought to the problem, try integrating with substitution and then differentiate your answer to see the process it went through to get to the final solution.

5. $\int x^3~e^{x^4}~{e^{e^{x^4}}}~dx$