Evaluate \( \int_{0}^{\infty} \frac{x^{8}\left(1-x^{6}\right)}{(1+x)^{24}} d x \)

To evaluate the integral \[ I = \int_{0}^{\infty} \frac{x^{8}(1-x^{6})}{(1+x)^{24}} \, dx, \] we can split it into two parts: \[ I = \int_{0}^{\infty} \frac{x^{8}}{(1+x)^{24}} \, dx - \int_{0}^{\infty} \frac{x^{14}}{(1+x)^{24}} \, dx. \] Let us denote these two integrals as \( I_1 \) and \( I_2 \): \[ I_1 = \int_{0}^{\infty} \frac{x^{8}}{(1+x)^{24}} \, dx, \] \[ I_2 = \int_{0}^{\infty} \frac{x^{14}}{(1+x)^{24}} \, dx. \] ### Step 1: Evaluate \( I_1 \) To evaluate \( I_1 \), we can use the beta function representation. The integral can be rewritten using the substitution \( x = \frac{t}{1-t} \), which gives \( dx = \frac{1}{(1-t)^2} dt \). The limits change from \( 0 \) to \( \infty \) to \( 0 \) to \( 1 \): \[ I_1 = \int_{0}^{1} \frac{\left(\frac{t}{1-t}\right)^{8}}{\left(1+\frac{t}{1-t}\right)^{24}} \cdot \frac{1}{(1-t)^2} \, dt. \] This simplifies to: \[ I_1 = \int_{0}^{1} \frac{t^{8}}{(1-t)^{8}} \cdot \frac{(1-t)^{24}}{(1-t+t)^{24}} \cdot \frac{1}{(1-t)^2} \, dt = \int_{0}^{1} \frac{t^{8}}{(1-t)^{10}} \cdot \frac{1}{(1)^{24}} \, dt. \] Thus, \[ I_1 = \int_{0}^{1} t^{8} (1-t)^{-10} \, dt. \] This integral can be evaluated using the beta function: \[ I_1 = B(9, 10) = \frac{\Gamma(9) \Gamma(10)}{\Gamma(19)}. \] Calculating the gamma functions: \[ \Gamma(9) = 8!, \quad \Gamma(10) = 9!, \quad \Gamma(19) = 18!. \] Thus, \[ I_1 = \frac{8! \cdot 9!}{18!}. \] Using the property \( n! = n \cdot (n-1)! \): \[ I_1 = \frac{8! \cdot 9 \cdot 8!}{18 \cdot 17 \cdots 10 \cdot 8!} = \frac{9 \cdot 8!}{18 \cdot 17 \cdots 10}. \] ### Step 2: Evaluate \( I_2 \) Similarly, for \( I_2 \): \[ I_2 = \int_{0}^{\infty} \frac{x^{14}}{(1+x)^{24}} \, dx. \] Using the same substitution \( x = \frac{t}{1-t} \): \[ I_2 = \int_{0}^{1} \frac{t^{14}}{(1-t)^{10}} \, dt = B(15, 10) = \frac{\Gamma(15) \Gamma(10)}{\Gamma(25)}. \] Calculating the gamma functions: \[ \Gamma(15) = 14!, \quad \Gamma(10) = 9!, \quad \Gamma(25) = 24!. \] Thus, \[ I_2 = \frac{14! \cdot 9!}{24!}. \] ### Step 3: Combine Results Now we can combine \( I_1 \) and \( I_2 \): \[ I = I_1 - I_2 = \frac{8! \cdot 9!}{18!} - \frac{14! \cdot 9!}{24!}. \] Factoring out \( 9! \): \[ I = 9! \left( \frac{8!}{