a) Đặt u = 1 + x4
\(\eqalign{
& \Rightarrow du = 4{x^3}dx \Rightarrow {x^3}dx = {{du} \over 4} \cr
& \int {{x^3}(1 + {x^4})dx = {1 \over 4}} \int {{u^3}du} = {{{u^4}} \over {16}} + c \cr&= {1 \over {16}}{(1 + {x^4})^4} + C \cr} \)
b) Ta có:
\(\int {\sin 2x.cosxdx = {1 \over 2}} \int {(\sin3x +\sin x)dx}\)
\(= - {1 \over 6} \cos 3x - {1 \over 2}\cos x + C\)
c) Ta có:
Đặt
\(\left\{ \matrix{ u = x \hfill \cr dv = {{dx} \over {{{\cos }^2}x}} \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ du = dx \hfill \cr v = \tan x \hfill \cr} \right.\)
Do đó:
\(\eqalign{
& \int {{x \over {{{\cos }^2}x}}} dx = x\tan x - \int {\tan xdx } \cr
& = x\tan x + \int {{{d(cosx)} \over {cosx}}} = x\tan x + \ln |cosx| + C \cr} \)
