∫_1^4▒〖ln⁡(√z+1)dz〗

Sure, let's solve the integral step by step:

1. First, let's simplify the integral a bit. We can rewrite √z as z^(1/2). So, the integral becomes ∫_1^4▒〖ln⁡(z^(1/2)+1)dz〗.

2. Now, let's use substitution to make the integral easier to solve. Let's set u = z^(1/2) + 1. Then, du = (1/2)z^(-1/2)dz.

3. We can solve for dz to get dz = 2z^(1/2)du. Substituting z = (u - 1)^2, we get dz = 2(u - 1)du.

4. Now, we can substitute u and dz into the integral. The limits of integration also change according to the substitution: when z = 1, u = 2 and when z = 4, u = 3. So, the integral becomes ∫_2^3▒〖ln⁡(u)(2(u - 1))du〗.

5. This integral can be solved using integration by parts, with v = ln(u) and dw = 2(u - 1)du. Then, dv = (1/u)du and w = u^2 - u.

6. According to the integration by parts formula, ∫v dw = vw - ∫w dv. Substituting v, w, dv, and dw, we get ∫_2^3▒〖ln⁡(u)(2(u - 1))du〗 = [ln(u)(u^2 - u)]_2^3 - ∫_2^3▒〖(u^2 - u)(1/u)du〗.

7. The first term can be evaluated directly to get [ln(3)(3^2 - 3) - ln(2)(2^2 - 2)] = 6ln(3) - 2ln(2).

8. The integral in the second term can be simplified to ∫_2^3▒〖(u - 1)du〗 = [u^2/2 - u]_2^3 = 3^2/2 - 3 - (2^2/2 - 2) = 1/2.

9. So, the original integral ∫_1^4▒〖ln⁡(√z+1)dz〗 = 6ln(3) - 2ln(2) - 1/2.