To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) given by:
\[
I_1 = \int_0^{\infty} \frac{x^2 \sqrt{x}}{(1+x)^6} \, dx
\]
\[
I_2 = \int_0^{\infty} \frac{x \sqrt{x}}{(1+x)^6} \, dx
\]
### Step 1: Simplifying the Integrals
First, we can rewrite \( \sqrt{x} \) as \( x^{1/2} \). Therefore, we can express the integrals as:
\[
I_1 = \int_0^{\infty} \frac{x^{5/2}}{(1+x)^6} \, dx
\]
\[
I_2 = \int_0^{\infty} \frac{x^{3/2}}{(1+x)^6} \, dx
\]
### Step 2: Using the Beta Function
To evaluate these integrals, we can use the substitution \( x = \frac{t}{1-t} \) which transforms the integral into a form involving the Beta function.
The Jacobian of this transformation is \( \frac{1}{(1-t)^2} \), and the limits change from \( 0 \) to \( 1 \) as \( x \) goes from \( 0 \) to \( \infty \).
### Step 3: Evaluating \( I_1 \)
For \( I_1 \):
\[
I_1 = \int_0^1 \frac{\left(\frac{t}{1-t}\right)^{5/2}}{(1+\frac{t}{1-t})^6} \cdot \frac{1}{(1-t)^2} \, dt
\]
This simplifies to:
\[
= \int_0^1 \frac{t^{5/2}}{(1-t)^{5/2}} \cdot \frac{(1-t)^6}{(1-t+t)^6} \cdot \frac{1}{(1-t)^2} \, dt
\]
\[
= \int_0^1 \frac{t^{5/2}}{(1-t)^{5/2}} \cdot \frac{(1-t)^4}{(1)^6} \, dt
\]
\[
= \int_0^1 t^{5/2} (1-t)^{4} \, dt
\]
This is a Beta function \( B\left(\frac{7}{2}, 5\right) \).
### Step 4: Evaluating \( I_2 \)
For \( I_2 \):
\[
I_2 = \int_0^1 \frac{\left(\frac{t}{1-t}\right)^{3/2}}{(1+\frac{t}{1-t})^6} \cdot \frac{1}{(1-t)^2} \, dt
\]
This simplifies to:
\[
= \int_0^1 \frac{t^{3/2}}{(1-t)^{3/2}} \cdot \frac{(1-t)^6}{(1-t+t)^6} \cdot \frac{1}{(1-t)^2} \, dt
\]
\[
= \int_0^1 t^{3/2} (1-t)^{4} \, dt
\]
This is a Beta function \( B\left(\frac{5}{2}, 5\right) \).
### Step 5: Relating \( I_1 \) and \( I_2 \)
Using the properties of the Beta function, we have:
\[
I_1 = B\left(\frac{7}{2}, 5\right) = \frac{\Gamma\left(\frac{7}{2}\right) \Gamma(5)}{\Gamma\left(\frac{7}{2}+5\right)}
\]
\[
I_2 = B\left(\frac{5}{2}, 5\right) = \frac{\Gamma\left(\frac{5}{2}\right) \Gamma(5)}{\Gamma\left(\frac{5}{2}+5\right)}
\]
To find the relationship between \( I_1 \) and \( I_2 \), we can use the property of the Gamma function and the fact that \( \Gamma(z+1) = z \Gamma(z) \).
### Final Result
After evaluating both integrals, we can find the relationship between \( I_1 \) and \( I_2 \) and conclude the problem.
