Consider the sequences of integrals: $\int \frac{x}{1+x} dx, \int \frac{x}{1+x^2}dx, \int \frac{x}{1+x^3}dx,\ldots,\int \frac{x}{1+x^n}dx$.
Question: What is the general formula of $\int \frac{x}{1+x^n} dx$ for $n>0$?
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Are these indefinite integrals, or is there a region of integration? – Ron Gordon Aug 02 '13 at 13:51
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1Is this any good for you: http://www.wolframalpha.com/input/?i=integrate+x%2F%281%2Bx%5En%29? – Ali Aug 02 '13 at 14:03
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$n$ is integer? – vesszabo Aug 02 '13 at 14:03
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2See this related problem. – Mhenni Benghorbal Aug 02 '13 at 14:11
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If you specify a particular region of $x$, then this can be easily solved by a contour integral. – Kunnysan Aug 02 '13 at 14:22
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It's not a complete answer, but too long for a comment. It gives the answer for $n=4m$, $m$ is positive integer. Denote $I_n:=\int\frac{x}{1+x^n}\,dx$. If $x<0$ then substituting $x:=-z$ we can reduce the problem for $x>0$. So we calculate $I_{4m}:=\int\frac{x}{1+x^{4m}}\,dx$. Substitute $x:=\sqrt{z},\,z>0$. Then $I_{4m}=\frac{1}{2}\int\frac{1}{1+z^{2m}}\,dz$. The result of the integral is (Demidovich, Mathematical Analysis, problem 1925) $$ -\frac{1}{2m}\sum_{k=1}^{m}\cos\frac{\pi(2k-1)}{2m}\ln\left(1-2x\cos\frac{2k-1}{2m}\pi+x^2 \right)$$ $$+\frac{1}{m}\sum_{k=1}^m\sin\frac{\pi(2k-1)}{2m}\arctan\frac{x-\cos\frac{2k-1}{2m}\pi}{\sin\frac{2k-1}{2m}\pi}. $$
vesszabo
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Possible first step: notice that $$ x\frac{1}{1+x^n}=x\sum_{k=0}^\infty (-1)^k x^{nk}=\sum_{k=0}^\infty (-1)^k x^{nk+1} $$
Ben Grossmann
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