pretty problem

$x = 5050 \frac{ \int_{0}^{1} ( 1- x^{50})^{100}}{ \int_{0}^{1} ( 1- x^{50})^{101}}$

applying by parts in the denominator and then putting the limits we get

$\frac{ \int_{0}^{1} ( 1- x^{50})^{100}}{ \int_{0}^{1} x^{50}( 1- x^{50})^{100}}$

$= \frac{ \int_{0}^{1} ( 1- x^{50})^{100}}{ \int_{0}^{1} (1 - ( 1 -x^{50}))( 1- x^{50})^{100}}$

$= \frac{ \int_{0}^{1} ( 1- x^{50})^{100}}{ \int_{0}^{1} ( 1 - x^{50})^{100} - \int_{0}^{1}( 1- x^{50})^{101}}$

$=\frac{1}{1 - \frac{ \int_{0}^{1} ( 1- x^{50})^{101}}{ \int_{0}^{1} ( 1- x^{50})^{100}}}$

$x = \frac{1}{1 - \frac{5050}{x}}$

thus $x = 5051$

$\displaystyle A=5050\frac{\displaystyle \int_0^1 (1-x^{50})^{100} dx }{\displaystyle \int_0^1 (1-x^{50})^{101} dx}$

In both integrals, use substitution $u=x^{50}$ to get :

$\displaystyle A=5050\frac{\displaystyle \int_0^1 u^{-\frac{49}{50}}(1-u)^{100} du }{\displaystyle \int_0^1 u^{-\frac{49}{50}}(1-u)^{101} du }$

(Recall Beta Function properties.)

$= 5050\frac{\displaystyle B(\frac{1}{50},101)}{\displaystyle B(\frac{1}{50},102)}= 5050\frac{\frac{\displaystyle \Gamma (\frac{1}{50})\Gamma (101)}{\displaystyle\Gamma (\frac{1}{50}+101)}}{\frac{\displaystyle\Gamma (\frac{1}{50})\Gamma (102)}{\displaystyle\Gamma (\frac{1}{50}+102)}}$

$\displaystyle = 5050\frac{\frac{\displaystyle\Gamma (\frac{1}{50})\Gamma (101)}{\displaystyle\Gamma (\frac{1}{50}+101)}}{\frac{\displaystyle\Gamma (\frac{1}{50})(101)\Gamma (101)}{\displaystyle (\frac{1}{50}+101)\Gamma (\frac{1}{50}+101)}} = \boxed{5051}$

( $\displaystyle \Gamma (x+1) = x\Gamma (x)$ )

Use beta function..!