Answer:
a. (24 ln 2 − 7) / 9
b. x tan x + ln|cos x| + C
Explanation:
a. ∫₁² x² ln x dx
Integrate by parts.
If u = ln x, then du = 1/x dx.
If dv = x² dx, then v = ⅓ x³.
∫ u dv = uv − ∫ v du
= (ln x) (⅓ x³) − ∫ (⅓ x³) (1/x dx)
= ⅓ x³ ln x − ∫ ⅓ x² dx
= ⅓ x³ ln x − ¹/₉ x³ + C
= ¹/₉ x³ (3 ln x − 1) + C
Evaluate between x=1 and x=2.
[¹/₉ 2³ (3 ln 2 − 1) + C] − [¹/₉ 1³ (3 ln 1 − 1) + C]
⁸/₉ (3 ln 2 − 1) + C + ¹/₉ − C
⁸/₉ (3 ln 2 − 1) + ¹/₉
⁸/₃ ln 2 − ⁸/₉ + ¹/₉
⁸/₃ ln 2 − ⁷/₉
(24 ln 2 − 7) / 9
b. ∫ x sec² x dx
Integrate by parts.
If u = x, then du = dx.
If dv = sec² x dx, then v = tan x.
∫ u dv = uv − ∫ v du
= x tan x − ∫ tan x dx
= x tan x + ∫ -sin x / cos x dx
= x tan x + ln|cos x| + C