∫_0^1▒(x dx)/(2+ √x)

To solve the integral ∫_0^1▒(x dx)/(2+ √x), we can use the substitution method. Here are the steps:

1. Let u = 2 + √x. Then, du = 1/(2√x) dx = dx/2√x.

2. We need to express dx in terms of du. To do this, we multiply both sides of the equation by 2√x to get dx = 2√x du.

3. Now, we substitute u and dx into the integral: ∫_0^1▒(x dx)/(2+ √x) = ∫_0^1▒(x * 2√x du)/u.

4. Simplify the integral to get ∫_0^1▒2x du/u.

5. Now, we need to change the limits of integration. When x = 0, u = 2 + √0 = 2. When x = 1, u = 2 + √1 = 3. So, the integral becomes ∫_2^3▒2x du/u.

6. The integral ∫_2^3▒2x du/u is not solvable because x is not a function of u. We made a mistake in step 4. The correct integral after substitution should be ∫_2^3▒(2u - 4) du/u.

7. Now, we can solve this integral: ∫_2^3▒(2u - 4) du/u = 2∫_2^3▒du - 4∫_2^3▒du/u.

8. The first integral is 2[u]_2^3 = 2(3 - 2) = 2. The second integral is -4[ln|u|]_2^3 = -4(ln3 - ln2).

9. So, the solution to the original integral is 2 - 4(ln3 - ln2) = 2 - 4ln(3/2).