Solve the integral: ∫1x2n+1dx

Question

Solve the integral:  ∫1x2n+1dx

Elaine Verrett · Accepted Answer

We have f(x)=1xn+1. Note that we can writef(x)=∏k=1n(x−xk)−1 (1)where xk=ei(2k−1)πn k=1,…,nWe can also express (1) asf(x)=∑k=1nak(x−xk)−1 (2)where ak=−xknNow, we can write∫1xn+1dx=−1n∑k=1nxklog⁡(x−xk)+Cwhich can be more explicitly written as∫1xn+1dx=−1n∑k=1n(12xkrlog⁡(x2−2xkrx+1)−xkiarctan⁡(x−xkrxki))+C′∣where xkr and xki are the real and imaginary parts of xk respectively, and are given byxr=Re(xk)=cos⁡((2k−1)πn)xki=Im(xk)=sin⁡((2k−1)πn)Note 1:The integral of ∫11+x2n is a special case for the development herein. Simply let n→2nNote 2:As requested, we will derive the form ak=−xkn. To that end, we use (2) and observe thatlimx→xl(

Tiefdruckot · Accepted Answer

It has no simple closed form, unless you also give some nice integration endpoints, such as ∫0∞. For your curiosity, you get x2F1(12n,1,1+12n,−x2n), where 2F1 is the hypergeometric function.

karton · Accepted Answer

First find x2n+1=0, then split into sum of fractions 1/(x+b) and 1/(x+b) and 1/(x2+bx+c) and integrate those. I seem to have forgotten what it&#039;s called in English. Not partial integration or integration by parts. Partial fraction decomposition maybe?

Solve the integral: \int\frac{1}{x^{2n}+1}dx

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