C. BAI LUYEN TAP Bai 1: Giai phuong trinh:
2064 Vi du 16: Tinh: a) J(l + tanx)2 e2x dx b) [xVd
Giai
a) J(l + tan)2e2xdx = f(l + tan2x + 2tanx)e2xdx = J(tanx.e2x
)dx =tanx.e2x + C
mm npHj ^g^r
b) Xet f(x) = (ax4 + bx3 + cx2 + dx + m)ex thi
f'(x) = (4ax3 + 3bx2 + 2cx + d)ex + (ax4 + bx3 + cx2 + dx + m)ex = [ax4 + (4a + b)x3 + (3b + c)x2 + (2c + d)x + d + m]ex
Ta co f ' ( x ) = x4ex <=> a = 1, 4a + b = 0, 3b + c = 0, 2c + d = 0, d + m = 0
oa = l , b = - 4 , c = 12, d = -24, m = 24.
Vay J x V d x = (x4 - 4x3 + 12x2 - 24x + 24)ex + C. DANG 2: PHUONG PHAP BIEN B 6 l BIEN S 6
Dang 1: Neu x = u(t) co dao ham lien tuc tren K thi: j f ( x ) d x = jf(u(t)).u'(t).dt
Dang 2: Neu t = v(x) co dao ham lien tuc tren K va co f(x)dx = g(t)dt thi: Jf(x)dx = Jg(t)dt
Chii y: - Su dung cac cong thuc mo rong kx voi k * 0.
M d rong cong thuc x thanh u kem theo du = u'.dx, luu y dau cong trir va he so nhan chia.
- Neu J f(x)dx = F(x) + C thi J f(u)du = F(u) + C
- Chon dat bien thich hop, bien u ma bieu thuc co san u' - Doi bien dai so: u = g(x), x = h(t), t = —, ...
x Dang , 1
thi dat t = x + %/x2
+ k
' V x2+ k
- Doi bien luong giac: Dang 9
1
9 thi dat x = a.tant x" -t- a"
Dang \la2 - x2 thi dat x = asint hay x = acost Dang V x2
+ a2
thi dat x = atant. Dang V x2 - a2 thi dat x = a
sin t
, x Bien doi theo goc phu t = tan—
2
Bac le voi sinx thi dat t = cosx, bac le vdi cosx thi dat t = sinx Dac biet d6i bien t = - x , t = TC - x, t = - x.
- Nguyen ham lien ket, de tinh I thi dat them J ma viec tinh I +J va I - J, I + kJ va I - mJ thuan loi hon, tu do suy ra I
V i d u 1: Tinh: a) f(2x + l )9d x c) j"x3(l + x4)3dx b) jx(3 - x)5dx d) f x Giai 18 dx
a) Doi bien u = 2x + 1 thi du = 2dx => dx = — du 2
f(2x + 2 )9d x = - f u9d u - . —. u1 0+ C = —( 2 x + l )1 0+ C J