Mathematical integration problems requiring techniques such as trigonometric identities, integration by parts, and substitution.

(a) ∫ tan²x dx = ∫ (sec²x - 1) dx = tan x - x + C

(b) Let u = ln x, dv = x⁻³ dx. Then du = (1/x) dx, v = -1/(2x²).
∫ (1/x³) ln x dx = -ln x / (2x²) - ∫ (-1/(2x²)) · (1/x) dx = -ln x / (2x²) + (1/2) ∫ x⁻³ dx = -ln x / (2x²) - 1/(4x²) + C

(c) Let u = 1 + eˣ, then du = eˣ dx, and e³ˣ = e²ˣ · eˣ = (u - 1)² · eˣ.
So ∫ e³ˣ / (1 + eˣ) dx = ∫ (u - 1)² / u du = ∫ (u² - 2u + 1)/u du = ∫ (u - 2 + 1/u) du = (1/2)u² - 2u + ln|u| + k
Substitute back: (1/2)(1 + eˣ)² - 2(1 + eˣ) + ln(1 + eˣ) + k = (1/2)(1 + 2eˣ + e²ˣ) - 2 - 2eˣ + ln(1 + eˣ) + k = (1/2)e²ˣ - eˣ + ln(1 + eˣ) + k'
(where k' absorbs constants)