Problems with calculating integrals

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:!) :smile: This question simply says calculate
Question
[integral from 0 to t]xe^-xdx

this is one of the questions from my calculus online course there are no similar examples to this in the given notebookk
i have taken the first year calculus but forgot somestuff, this was 3 years ago, this is why i really need help!
Hints is fine as long of course they lead me to the solution.
This is what i thought i can do , rearrange the equation to somthing like x/e^x dx but now what if i use substitution say u= e^x then du= e^xdx
now the new limits would go like this whwen x= 0 then u= 1 and when the upper limit x=t then u=e^t i think so far so good, i hope someone can confirm the aligibility of this ??!
Then :
[integral from 0 to t]x/e^xdx
=[integral from 1 to e^t]x/u[1/e^x]du
now i don't know what to do:rolleyes: :rolleyes:


Another question:
is there a specific method for calculation integrals like this?
[integral from 2 to 3 ] 1/[x-1][x-4]dx
also [integral from 0 to 3] 1/([9-x^2]^1/2) dx
Last but not least is there a website anyoone suggests that has examples of such problems?Thank you for the help
 
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For your first, use integration by parts.
For your second, use partial fractions decomposition.
For your third, use the substitution x=3\sin(u)
 
Tabular also works for the first one, if you're familiar with this method. It uses by parts, but it's less time consuming.
 
x/e^x= xe^(-x) doesn't it? Looks like a good candidate for integration by parts.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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