Evaluating the Integral $\int_{0}^{\infty} \frac{x^{49}}{(1+x)^{51}} dx$ I tried evaluating the integral $\displaystyle \int_{0}^{\infty} \dfrac{x^{49}}{(1+x)^{51}}dx$
but I wasn't able to get the result.
Following is the way by which I did it-
$$I=\displaystyle \int_{0}^{\infty} \dfrac{x^{49}}{(1+x)^{51}}dx$$
$$\implies I=\int_{0}^{\infty} x^{49}(1+x)^{-51}dx$$
Further, I tried Integration by parts but it didn't worked. Can anyone tell that how this integral can be evaluated.

Here is an alternative method.
First of all, for each integer $n>1$, we have $\int_1^\infty x^{-n}dx=\frac1{n-1}$.
Thus, using the binomial theorem,
\begin{align}
\int_0^\infty\frac{x^{49}}{(1+x)^{51}}dx&=\int_1^\infty\frac{(x-1)^{49}}{x^{51}}dx\\
&=\int_1^\infty\frac1{x^{51}}\sum_{n=0}^{49}(-1)^{n-1}{49\choose n}x^ndx\\
&=\sum_{n=0}^{49}(-1)^{n-1}{49\choose n}\int_1^\infty x^{n-51}dx\\
&=\sum_{n=0}^{49}(-1)^{n-1}{49\choose n}\frac1{50-n}\\
&=\frac1{50}\sum_{n=0}^{49}(-1)^{n-1}{50\choose n}\\
&=-\frac1{50}\left((1+(-1))^{50}-{50\choose50}(-1)^{50}\right)\\
&=\frac1{50}.
\end{align}
