Compute the indefinite integral $\displaystyle{\int{\frac{t^{n-1}-1}{(t+1)^{n+1}}dt}}$.
I tried to solve it by using binomial expansion for $(t+1)^{n+1}$ and then writing $\displaystyle{\frac{t^{n-1}-1}{(t+1)^{n+1}} = \frac{A_1}{t+1} + \frac{A_2}{(t+1)^2}}+...+\frac{A_{n+1}}{(t+1)^{n+1}}$, but it does take a lot of work. Are there any other methods?
First, split the fraction.
$$\int\dfrac{t^{n-1}-1}{(t+1)^{n+1}}dt=\int\dfrac{t^{n-1}dt}{(t+1)^{n+1}}-\int\dfrac{dt}{(t+1)^{n+1}}$$
The second integral should be straightforward. As for the first, multiply top and bottom by $t^2$.
$$\int\dfrac{t^{n+1}dt}{(t+1)^{n+1}t^2}=\int\frac1{t^2}\left(\frac{t}{t+1}\right)^{n+1}dt=\int\frac1{t^2}\left(\dfrac1{1+t^{-1}}\right)^{n+1}dt$$
Now just substitute $u=1+t^{-1}$