The problem asks to evaluate the definite integral $\int_{0}^{1} x\sqrt{x^2+1} dx$.

1. Problem Description

The problem asks to evaluate the definite integral $\int_{0}^{1} x\sqrt{x^2+1} dx$.

2. Solution Steps

To evaluate the integral, we can use u-substitution.

Let $u = x^2 + 1$. Then, $\frac{du}{dx} = 2x$, so $du = 2x \, dx$.

Therefore, $x \, dx = \frac{1}{2} du$.

Also, when $x = 0$, $u = 0^2 + 1 = 1$.

When $x = 1$, $u = 1^2 + 1 = 2$.

Thus, the integral becomes:

$\int_{0}^{1} x\sqrt{x^2+1} \, dx = \int_{1}^{2} \sqrt{u} \cdot \frac{1}{2} \, du = \frac{1}{2} \int_{1}^{2} u^{\frac{1}{2}} \, du$

Now, we can find the antiderivative of $u^{\frac{1}{2}}$:

$\int u^{\frac{1}{2}} \, du = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C = \frac{2}{3} u^{\frac{3}{2}} + C$

So, the integral becomes:

$\frac{1}{2} \int_{1}^{2} u^{\frac{1}{2}} \, du = \frac{1}{2} \left[ \frac{2}{3} u^{\frac{3}{2}} \right]_{1}^{2} = \frac{1}{3} \left[ u^{\frac{3}{2}} \right]_{1}^{2}$

Evaluating the definite integral:

$\frac{1}{3} \left( 2^{\frac{3}{2}} - 1^{\frac{3}{2}} \right) = \frac{1}{3} (2\sqrt{2} - 1)$

3. Final Answer

$\frac{2\sqrt{2}-1}{3}$