Help evaluating long definite integral. First, thanks for your time and input. In some research, I evaluate the following integral 
$$\int_0^1 x (n-1) n v^{-2n}(v-x)\lambda^{-n}(v-2x+\lambda v)(\lambda v-x)(x(v+\lambda v-x))^{n-2} dx$$ where v>0  $\lambda$>1 and n>2. using Mathematica and the solution is very ugly (including hypergeometric functions). 
I was wondering if anyone could help advise me on how to start to evaluate this integral by hand.
Thanks!
 A: Well, simply rewrite your function. Here's how to do it by hand (as you asked):
First let's write:
$$\displaystyle I=\int_0^1 x (n-1) n v^{-2n}(v-x)\lambda^{-n}(v-2x+\lambda v)(\lambda v-x)(x(v+\lambda v-x))^{n-2} dx=\int_0^1f(x)dx$$
I'll just get rid of all the constant terms and put them on the side:
$f(x)=\left((n-1)nv^{-2n}\lambda^{-n}\right)\left(x(v-x)(v-2x+\lambda v)(\lambda v-x)x^{n-2}(v+\lambda v-x)^{n-2}\right)$
Now let's introduce some notations:
$$\left\{\begin{array}{cc}\left((n-1)nv^{-2n}\lambda^{-n}\right)=A \\ v+\lambda v=B\end{array}\right.$$
Replace in $I$:
$$\displaystyle I=\int_0^1 Ax(v-x)(B-2x)(\lambda v-x)x^{n-2}(B-x)^{n-2}dx$$
Now get rid of this constant term $A$ and call this new function in the integral $g$:
$$\displaystyle I=A\int_0^1 (v-x)(B-2x)(\lambda v-x)x^{n-1}(B-x)^{n-2}dx=A\int_0^1g(x)dx$$
Expand $g$:
$$g(x)=(v-x)(B-2x)(\lambda v-x)x^{n-1}(B-x)^{n-2}$$
$$=(vB-(B+2v)x+2x^2)(\lambda v-x)x^{n-1}(B-x)^{n-2}$$
$$=((v^2B\lambda)-x(vB+(B+2v)\lambda v)+x^2(2\lambda v+B+2v)-2x^3)x^{n-1}(B-x)^{n-2}$$
Add some more notations (just because I don't want to carry all these constants around):
$$\left\{\begin{array}{cc}(v^2B\lambda)=C_1 \\ (vB+(B+2v)\lambda v)=C_2 \\ (2\lambda v+B+2v)=C_3 \end{array}\right.$$
Your function looks a lot better now:
$$g(x)=\left(C_1x^{n-1}-C_2x^{n}+C_3x^{n+1}-2x^{n+2}\right)(B-x)^{n-2}$$
The "hard" part is the $(B-x)^{n-2}$, but not that hard actually, because you can write it this way:
$$\displaystyle(B-x)^{n-2}=\sum_{k=0}^{n-2}{n-2 \choose k}B^{n-2-k}(-x)^k=\sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}x^k$$
So $g$ becomes:
$$g(x)=\left(C_1x^{n-1}-C_2x^{n}+C_3x^{n+1}-2x^{n+2}\right)\sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}x^k$$
$$g(x)=\sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}\left(C_1x^{n-1}-C_2x^{n}+C_3x^{n+1}-2x^{n+2}\right)x^k$$
$$g(x)=\sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}\left(C_1x^{k+n-1}-C_2x^{k+n}+C_3x^{k+n+1}-2x^{k+n+2}\right)$$
Let's go back to the integral:
$$\displaystyle I=A\int_0^1 g(x)dx$$
$$\displaystyle I=A\int_0^1 \sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}\left(C_1x^{k+n-1}-C_2x^{k+n}+C_3x^{k+n+1}-2x^{k+n+2}\right) dx$$
$$\displaystyle I=A\sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}\int_0^1 \left(C_1x^{k+n-1}-C_2x^{k+n}+C_3x^{k+n+1}-2x^{k+n+2}\right) dx$$
$$\displaystyle I=A\sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}\times J_k$$
where $\displaystyle J_k=\int_0^1 \left(C_1x^{k+n-1}-C_2x^{k+n}+C_3x^{k+n+1}-2x^{k+n+2}\right) dx$
Now you have a simple polynomial to integrate:
$$\displaystyle J_k=\left[\frac{C_1}{k+n}x^{k+n}-\frac{C_2}{k+n+1}x^{k+n+1}+\frac{C_3}{k+n+2}x^{k+n+2}-\frac{2}{k+n+3}x^{k+n+3}\right]_0^1$$
$$\displaystyle J_k=\frac{C_1}{k+n}-\frac{C_2}{k+n+1}+\frac{C_3}{k+n+2}-\frac{2}{k+n+3}$$
And thus you have:
$$\displaystyle I=A\sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}\left(\frac{C_1}{k+n}-\frac{C_2}{k+n+1}+\frac{C_3}{k+n+2}-\frac{2}{k+n+3}\right)$$
You can replace your values $A$,$B$,$C_1$,$C_2$ and $C_3$ and get your result (which I won't write here).
I admit it's still pretty ugly but at least you only have simple fractions and sums (and binomial terms). But with a function like yours you can't just expect a nice and clean solution ;)
