Answer:
a) e^x (3 - e^x)^4 dx
Using the formula [ f'(x)[f(x)"]dx =[ [f(x)]^n+1 / n+1] + c, where n = 1,
we have:
e^x (3 - e^x)^4 dx = (e^x)^5 (3 - e^x)^4 / 5 + c
= (e^5x - 4e^4x + 6e^3x - 4e^2x + e^x) / 5 + c
b) 3e^2x √(1 + e²x) dx
Using the formula [ f'(x)[f(x)"]dx =[ [f(x)]^n+1 / n+1] + c, where n = 1,
we have:
3e^2x √(1 + e²x) dx = (3e^2x)^2 * (1 + e²x)^(3/2) / 2 + c
= (9e^4x + 3e^2x) / 2 + c
c) 3e^-2x / (1 + e^-2x)^3 dx
Using the formula [ f'(x)[f(x)"]dx =[ [f(x)]^n+1 / n+1] + c, where n = 1,
we have:
3e^-2x / (1 + e^-2x)^3 dx = -(3e^-2x)^2 / (1 + e^-2x)^2 + c
= -(9e^-4x) / (e^-4x + 2e^-2x + 1) + c
d) 4 cos 2x sin³ 2x dx
Using the formula [ f'(x)[f(x)"]dx =[ [f(x)]^n+1 / n+1] + c, where n = 1,
we have:
4 cos 2x sin³ 2x dx = -4 cos 2x (sin 2x)^4 / 4 + c
= -(cos 2x) (1 - cos 4x)^2 / 2 + c
e) sec² 3x tan³ 3x dx
Using the formula [ f'(x)[f(x)"]dx =[ [f(x)]^n+1 / n+1] + c, where n = 1,
we have:
sec² 3x tan³ 3x dx = -sec² 3x (tan 3x)^4 / 4 + c
= -sec² 3x (sec² 3x - 1)^2 / 4 + c
f) 2+tan ² x / cos² x dx
Using the formula [ f'(x)[f(x)"]dx =[ [f(x)]^n+1 / n+1] + c, where n = 1,
we have:
2+tan ² x / cos² x dx = ln|sec x| + c
It's worth noting that all these integrals are indefinite, which means that the constant c is arbitrary, and the actual antiderivative depends on the problem context.
Explanation: