Calculating the integral of x^n e^(2x) from 0 to 1

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Hey, I got a couple of problems on integration that I can't seem to figure out.

1. Suppose I_n = \int_{0}^{1} x^n e^{2x} dx Evaluate I_n in terms of I_{n-1} for any natural number n


2. Suppose I_n = \int_{1}^{e} \left[\ln x\right]^n dx Evaluate I_n in terms of I_{n-1} for any natural number n

Not sure what to do with these. Do I need to integrate I_{n-1}? What do I put into I_n after I integrate I_{n-1}??
 
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With

I_n = \int_{0}^{1} x^n e^{2x} dx

then

I_{n-1} = \int_{0}^{1} x^{n-1} e^{2x} dx

Use integration by parts to get a relation between I_n and I_n-1.
 
Yup, I did that and I got I_n = \frac{1}{2}x^ne^{2x} - \frac{n}{2}I_{n-1}

For the second question, I let u = ln x, but what would I let dv/dx equal to?
 
It's a bit unclear to me what you mean with dv/dx, but do you mean identifying the factors for integration by parts?

Remember that one can integrate ln(x)dx by looking at it as ln(x).1dx and taking u = ln(x) and v = 1, dv = dx.
 
Yup! I_n = x[\ln x]^n - nI_{n-1}
 
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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