Find the value of `int_0^1x(1-x)^ndx`

Explanation

To solve the integral \( I = \int_0^1 x(1-x)^n \, dx \), we can follow these steps:

### Step 1: Define the integral
Let
\[ I = \int_0^1 x(1-x)^n \, dx. \]

### Step 2: Substitute \( x \) with \( 1-x \)
Now, we will perform a substitution by replacing \( x \) with \( 1-x \):
\[ I = \int_0^1 (1-x)(1-(1-x))^n \, dx = \int_0^1 (1-x)x^n \, dx. \]

### Step 3: Simplify the integral
The integral can be expressed as:
\[ I = \int_0^1 (1-x)x^n \, dx = \int_0^1 (x^n - x^{n+1}) \, dx. \]

### Step 4: Separate the integrals
Now we can separate the integral:
\[ I = \int_0^1 x^n \, dx - \int_0^1 x^{n+1} \, dx. \]

### Step 5: Compute the integrals
Using the formula for the integral of \( x^k \):
\[ \int x^k \, dx = \frac{x^{k+1}}{k+1} + C, \]
we calculate:
\[ \int_0^1 x^n \, dx = \left[ \frac{x^{n+1}}{n+1} \right]_0^1 = \frac{1^{n+1}}{n+1} - 0 = \frac{1}{n+1}, \]
and
\[ \int_0^1 x^{n+1} \, dx = \left[ \frac{x^{n+2}}{n+2} \right]_0^1 = \frac{1^{n+2}}{n+2} - 0 = \frac{1}{n+2}. \]

### Step 6: Substitute back into the equation
Substituting these results back into our expression for \( I \):
\[ I = \frac{1}{n+1} - \frac{1}{n+2}. \]

### Step 7: Simplify the expression
Finding a common denominator:
\[ I = \frac{(n+2) - (n+1)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)}. \]

### Final Result
Thus, the value of the integral is:
\[ I = \frac{1}{(n+1)(n+2)}. \]

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