Evaluate $$\int_0^1 t^m(1+t)^n dt$$
My try:
Let $$I(m,n)= \int_0^1 t^m(1+t)^n dt=\int_0^1 t^{m-1} (1+t-1)(1+t)^n dt=I(m-1,n+1)- I(m-1,n)$$
And by using Integration by parts we get $$I(m,n)=\frac {2^{n+1}}{n+1}-\frac {m}{n+1} I(m-1,n+1)$$
From these two equations we get $$I(m,n)=\frac {2^{n+1}}{m+n+1} - \frac {m}{m+n+1} I(m-1,n)$$
Now I was only asked to find the recurrence but I am wondering if there could be any closed form solution to it. So in the process of finding the closed form I found out that $$I(m,n)=\frac {2^{n+1}}{m+n+1} - \frac {m\cdot 2^{n+1}}{(m+n+1)(m+n)}+\frac {m(m-1)\cdot 2^{n+1}}{(m+n+1)(m+n)(m+n-1)}-\cdots\cdots +\frac {(-1)^{m-1}m!\cdot 2^{n+1}}{(m+n+1)(m+n)(m+n-1)\cdots (n+2)}+\frac {(-1)^{m}m!\cdot 2^{n+1}}{(m+n+1)(m+n)(m+n-1)\cdots (n+2)(n+1)}+ \frac {(-1)^m \cdot m!}{ (m+n+1)(m+n)(m+n-1)\cdots (n+1)}$$
The series seems to follow some pattern But I am not able to simplify this summation. I tried writing the general term but couldn't find a proper one.
Any help would be greatly appreciated
There are a few options, none of them especially neat. (I also recommend you to double-check these calculations.) You can use the substitution $u=\dfrac{t}{1+t^2}$ to write the integral as an incomplete Beta function. It's easiest to do this by writing $t=\tan^2 x,\,u=\sin^2 x$, viz.$$I(m,\,n)=2\int_0^{\pi/4}\sin^{2m+1}x\cos^{-2m-2n-3}x dx=\operatorname{B}(\frac{1}{2};\,m+1,\,-m-n-1).$$
For $n$ a non-negative integer you can use$$I(m,\,n)=\sum_{k=0}^n\frac{\binom{n}{k}}{m+k+1}.$$For $m$ a non-negative integer, you can define $f_n(a):=\int_0^1 (1+at)^n\text{d}t=\dfrac{(1+a)^{n+1}-1}{a(n+1)}$ so $I(m,\,n)=\dfrac{f_{m+n}^{(m)}(1)}{\prod_{i=1}^m (n+i)}$.