The integral of f(x) is (1/36) ln|u| + 36 + C.
Here are the integrals of the given functions:
f(x) = x(x + 4)^2
We can expand the polynomial to get:
f(x) = x(x^2 + 8x + 16)
Then, we can integrate each term separately:
∫ x(x^2 + 8x + 16) dx = ∫ x^3 dx + 8∫ x^2 dx + 16∫ x dx
Using the power rule of integration, we get:
= (1/4)x^4 + (8/3)x^3 + 8x^2 + C
Therefore, the integral of f(x) is (1/4)x^4 + (8/3)x^3 + 8x^2 + C.
f(x) = e^(k+ae)
We can treat k as a constant and integrate with respect to x:
∫ e^(k+ae) dx = (1/a) e^(k+ae) + C
Therefore, the integral of f(x) is (1/a) e^(k+ae) + C.
f(x) = 5 √x
We can rewrite the function as:
f(x) = 5x^(1/2)
Then, we can integrate using the power rule:
∫ 5x(1/2) dx = (5/4) (2/3) x^(3/2) + C
= (5/6) x^(3/2) + C
Therefore, the integral of f(x) is (5/6) x^(3/2) + C.
f(x) = 6/(5^(4x))
We can use the substitution u = 5^4x, du/dx = 20(5^3)x^3, and dx = du/(20(5^3)x^3) to rewrite the integral as:
∫ 6/5^(4x) dx = (6/5) ∫ 1/u^4 du
= (6/5) (1/49) u^(-4) + C
= (6/5) (1/49) (5^4x)^{-4} + C
Therefore, the integral of f(x) is (6/5) (1/49) (5^4x)^{-4} + C.
f(x) = 1/x^2 + 36
We can use the substitution u = 1/x, du/dx = -1/x^2, and dx = du/(1/x) to rewrite the integral as:
∫ 1/x^2 + 36 dx = (1/36) ∫ du - 36 ∫ du/x^2
= (1/36) ln|u| - 36 (-1/u) + C
= (1/36) ln|u| + 36 + C
Therefore, the integral of f(x) is (1/36) ln|u| + 36 + C.