We are asked to evaluate the definite integral $\int_0^1 x(1+x)^{2022} dx$.

1. Problem Description

We are asked to evaluate the definite integral $\int_0^1 x(1+x)^{2022} dx$.

2. Solution Steps

Let $u = 1+x$. Then $x = u-1$ and $dx = du$.

When $x = 0$, $u = 1+0 = 1$.

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

Thus, the integral becomes

$\int_1^2 (u-1)u^{2022} du = \int_1^2 (u^{2023} - u^{2022}) du$.

We integrate using the power rule:

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$.

Thus,

$\int_1^2 (u^{2023} - u^{2022}) du = \left[\frac{u^{2024}}{2024} - \frac{u^{2023}}{2023}\right]_1^2 = \left(\frac{2^{2024}}{2024} - \frac{2^{2023}}{2023}\right) - \left(\frac{1^{2024}}{2024} - \frac{1^{2023}}{2023}\right) = \frac{2^{2024}}{2024} - \frac{2^{2023}}{2023} - \frac{1}{2024} + \frac{1}{2023}$.

We can simplify this expression:

$\frac{2^{2024}}{2024} - \frac{2^{2023}}{2023} - \frac{1}{2024} + \frac{1}{2023} = \frac{2^{2023} \cdot 2}{2024} - \frac{2^{2023}}{2023} - \frac{1}{2024} + \frac{1}{2023} = 2^{2023} \left(\frac{2}{2024} - \frac{1}{2023}\right) - \frac{1}{2024} + \frac{1}{2023} = 2^{2023} \left(\frac{1}{1012} - \frac{1}{2023}\right) + \left(\frac{1}{2023} - \frac{1}{2024}\right) = 2^{2023} \left(\frac{2023 - 1012}{1012 \cdot 2023}\right) + \frac{2024 - 2023}{2023 \cdot 2024} = 2^{2023} \left(\frac{1011}{1012 \cdot 2023}\right) + \frac{1}{2023 \cdot 2024} = \frac{2^{2023} \cdot 1011}{1012 \cdot 2023} + \frac{1}{2023 \cdot 2024}$.

Combining fractions, we have

$\frac{2^{2023} \cdot 1011 \cdot 2024 + 1012}{1012 \cdot 2023 \cdot 2024} = \frac{2^{2023} \cdot (1012 - 1) \cdot 2024 + 1012}{1012 \cdot 2023 \cdot 2024} = \frac{2^{2023} \cdot 1012 \cdot 2024 - 2^{2023} \cdot 2024 + 1012}{1012 \cdot 2023 \cdot 2024} = \frac{2^{2023} \cdot 1012 \cdot 2024 + 1012 - 2^{2023} \cdot 2024}{1012 \cdot 2023 \cdot 2024} = \frac{1012 (2^{2023} \cdot 2024 + 1) - 2^{2023} \cdot 2024}{1012 \cdot 2023 \cdot 2024}$.

$\frac{2^{2024}}{2024} - \frac{2^{2023}}{2023} - \frac{1}{2024} + \frac{1}{2023} = \frac{2023(2^{2024}-1) - 2024(2^{2023}-1)}{2023 \cdot 2024} = \frac{2023 \cdot 2^{2024} - 2023 - 2024 \cdot 2^{2023} + 2024}{2023 \cdot 2024} = \frac{2^{2023}(2023 \cdot 2 - 2024) + 1}{2023 \cdot 2024} = \frac{2^{2023}(4046 - 2024) + 1}{2023 \cdot 2024} = \frac{2022 \cdot 2^{2023} + 1}{2023 \cdot 2024}$.

3. Final Answer

$\frac{2022 \cdot 2^{2023} + 1}{2023 \cdot 2024}$