If the integral $I = \int \frac{x^2}{\sqrt{1+x}} dx = \frac{1}{\alpha} (1+x)^\alpha - \frac{8\alpha}{15} (1+x)^{\alpha-1} + 2(1+x)^{\alpha-2} + C$, $C$ is constant of integration, then the value of $\alpha$ is:

To solve the integral, we use the method of substitution.

Let $1 + x = t$. Then $x = t - 1$ and $dx = dt$.

Substituting these into the original expression:

$I = \int \frac{(t-1)^2}{\sqrt{t}} dt$

Expanding the numerator $(t-1)^2 = t^2 - 2t + 1$:

$I = \int \left( \frac{t^2}{t^{1/2}} - \frac{2t}{t^{1/2}} + \frac{1}{t^{1/2}} \right) dt$

Simplify the exponents:

$I = \int (t^{3/2} - 2t^{1/2} + t^{-1/2}) dt$

Apply the power rule $\int t^n dt = \frac{t^{n+1}}{n+1}$:

$I = \frac{t^{5/2}}{5/2} - 2 \cdot \frac{t^{3/2}}{3/2} + \frac{t^{1/2}}{1/2} + C$

$I = \frac{2}{5}t^{5/2} - \frac{4}{3}t^{3/2} + 2t^{1/2} + C$

Replace $t$ with $1 + x$:

$I = \frac{2}{5}(1+x)^{5/2} - \frac{4}{3}(1+x)^{3/2} + 2(1+x)^{1/2} + C$

Comparing this to the given form:

$I = \frac{1}{\alpha} (1+x)^\alpha - \frac{8\alpha}{15} (1+x)^{\alpha-1} + 2(1+x)^{\alpha-2} + C$

By comparing the first term $\frac{2}{5}(1+x)^{5/2}$ with $\frac{1}{\alpha}(1+x)^\alpha$, we find:

$\alpha = \frac{5}{2}$