This is an old calculus question:
Integrate $$\int x^{-2n-2}(1-x)^n(1-cx)^ndx$$ it comes from Edwards Treatise on Integral Calculus Vol I, pg. 65 and is designated as [Oxford I.P. 1917].
I guess my question is what do they have in mind as a solution to this ? Are they expecting a closed form ?
My solution is to expand the $1-cx$ factor to get $(-1)^{k}\binom{n}{k} \int\frac{x^k(1-x)^n}{x^{2n+2}}dx$ as the coefficient of $c^k$. This in turn can be integrated by the binomial theorem as $\sum_{l=0}^n(-1)^{k+l}\binom{n}{k}\binom{n}{l}\frac{1}{2n-l-k+1}x^{-2n+k+l-1}$. This almost seems too simple minded. But maybe it was what they intended. Any opinions ?
This integral can be written as a simple summation. To simplify, we change $y=1/x$ to write \begin{align} I_n=&\int x^{-2n-2}(1-x)^n(1-cx)^n\,dx\\ &=-\int (y-1)^n(y-c)^n\,dy \end{align} and to have a symmetric expression, changing $y=z+(c+1)/2$ gives \begin{align} I_n&=-\int \left( z^2-\left( \frac{c-1}{2} \right)^2 \right)^n\,dz \end{align} Now, expanding the polynomial, \begin{align} I_n&=-\sum_{k=0}^n\binom{n}{k}\frac{(-1)^{n-k}}{2k+1}\left( \frac{c-1}{2} \right)^{2n-2k}z^{2k+1}+C\\ &=(-1)^{n+1}\sum_{k=0}^n\binom{n}{k}\frac{(-1)^{k}}{2k+1}\left( \frac{c-1}{2} \right)^{2n-2k}\left(1-\frac{(c+1)x}{2} \right)^{2k+1}x^{-2k-1}+C \end{align}