Đang tải... (xem toàn văn)
MËT SÈ K THUT "CHUÈI" M TA TH×ÍNG GP KHI I TM NGUYN HM HOC TNH TCH PH N. Th½ dö 1 : T¼m nguy¶n h m A 1 = Z (x + 1) (dxx + 2) Ta gi£ sû r¬ng : 1 (x + 1) (x + 2) = α x + 1 + β x + 2 Ta i t¼m 2 h» sè α , β theo hai c¡ch nh÷ sau : C¡ch 1 : α = lim x →− 1 x + 1 (x + 1) (x + 2) = limx →− 1 x + 21 = −1 + 21 = 1 β = lim x →− 2 x + 2 (x + 1) (x + 2) = limx →− 2 x + 11 = −2 + 11 = −1 C¡ch 2 : Cho x = 0 ta câ : 1 (0 + 1) (0 + 2) = α 0 + 1 + β 0 + 2 ⇔ 12 = α + 12 β x = 1 ta câ : 1 (1 + 1) (1 + 2) = α 1 + 1 + β 1 + 2 ⇔ 16 = 12 α + 13 β Do â m ta suy ra 2 h» sè α , β b¬ng c¡ch i gi£i h» : α12 α++12 β13β==12 16 ⇔ ( αβ = 1−1 C¡ch 3 : Ta gi£ sû r¬ng : 1 (x + 1) (x + 2) = α x + 1 + β x + 2 ⇔ 1 (x + 1) (x + 2) = α (x + 2) + β (x + 1) (x + 1) (x + 2) = x (α + β) + 2α + β (x + 1) (x + 2) C¥n b¬ng c¡c h» sè 2 v¸ m ta câ h» : ( α2α++ββ= 0= 1 ⇔ ( αβ = 1−1 Do â m ta suy ra : 1 (x + 1) (x + 2) = 1 x + 1 − 1 x + 2 Vªy : Z (x + 1) (dxx + 2) = Z xdx+ 1 − Z xdx+ 2 = ln |x + 1| − ln |x + 2| + c = ln x + 1+ 2 + c Th½ dö 2 : T¼m nguy¶n h m A 2 = Z x (x −x2) (+ 2x + 5) dx Ta gi£ sû r¬ng : x + 2 x (x − 2) (x + 5) = αx + β x − 2 + χ x + 5 C¡ch 1 : α = lim x → 0 x (x + 2) x (x − 2) (x + 5) = limx → 0 (x −x2) (+ 2x + 5) = (0 − 0 + 22) (0 + 5) = − 15 β = lim x → 2 (x − 2) (x + 2) x (x − 2) (x + 5) = limx → 2 x (xx+ 2+ 5) = 2 (2 + 5)2 + 2 = 27 χ = lim x →− 5 (x + 5) (x + 2) x (x − 2) (x + 5) = limx →− 5 x (xx+ 2− 2) = −5 (−−5 + 25 − 2) = − 353 C¡ch 2 : Cho x = −1 ta câ : −1 + 2 −1 (−1 − 2) (−1 + 5) = α−1 + β −1 − 2 + χ −1 + 5 ⇔ −α − 13 β + 1 4 χ = 1 12 x = 1 ta câ : 1 + 2 1 (1 − 2) (1 + 5) = α1 + β 1 − 2 + χ 1 + 5 ⇔ α − β + 1 6 χ = − 12
A 1 = dx (x + 1) (x + 2) 1 (x + 1) (x + 2) = α x + 1 + β x + 2 α , β α = lim x→−1 x + 1 (x + 1) (x + 2) = lim x→−1 1 x + 2 = 1 −1 + 2 = 1 β = lim x→−2 x + 2 (x + 1) (x + 2) = lim x→−2 1 x + 1 = 1 −2 + 1 = −1 x = 0 1 (0 + 1) (0 + 2) = α 0 + 1 + β 0 + 2 ⇔ 1 2 = α + 1 2 β x = 1 1 (1 + 1) (1 + 2) = α 1 + 1 + β 1 + 2 ⇔ 1 6 = 1 2 α + 1 3 β α , β α + 1 2 β = 1 2 1 2 α + 1 3 β = 1 6 ⇔ α = 1 β = −1 1 (x + 1) (x + 2) = α x + 1 + β x + 2 ⇔ 1 (x + 1) (x + 2) = α (x + 2) + β (x + 1) (x + 1) (x + 2) = x (α + β) + 2α + β (x + 1) (x + 2) α + β = 0 2α + β = 1 ⇔ α = 1 β = −1 1 (x + 1) (x + 2) = 1 x + 1 − 1 x + 2 dx (x + 1) (x + 2) = dx x + 1 − dx x + 2 = ln |x + 1| − ln |x + 2| + c = ln x + 1 x + 2 + c A 2 = x + 2 x (x − 2) (x + 5) dx x + 2 x (x − 2) (x + 5) = α x + β x − 2 + χ x + 5 α = lim x→0 x (x + 2) x (x − 2) (x + 5) = lim x→0 x + 2 (x − 2) (x + 5) = 0 + 2 (0 − 2) (0 + 5) = − 1 5 β = lim x→2 (x − 2) (x + 2) x (x − 2) (x + 5) = lim x→2 x + 2 x (x + 5) = 2 + 2 2 (2 + 5) = 2 7 χ = lim x→−5 (x + 5) (x + 2) x (x − 2) (x + 5) = lim x→−5 x + 2 x (x − 2) = −5 + 2 −5 (−5 − 2) = − 3 35 x = −1 −1 + 2 −1 (−1 − 2) (−1 + 5) = α −1 + β −1 − 2 + χ −1 + 5 ⇔ −α − 1 3 β + 1 4 χ = 1 12 x = 1 1 + 2 1 (1 − 2) (1 + 5) = α 1 + β 1 − 2 + χ 1 + 5 ⇔ α − β + 1 6 χ = − 1 2 x = 3 3 + 2 3 (3 − 2) (3 + 5) = α 3 + β 3 − 2 + χ 3 + 5 ⇔ 1 3 α + β + 1 8 χ = 5 24 α , β , χ −α − 1 3 β + 1 4 χ = 1 12 α − β + 1 6 χ = − 1 2 1 3 α + β + 1 8 χ = 5 24 ⇔ α = − 1 5 β = 2 7 χ = − 3 35 x + 2 x (x − 2) (x + 5) = α x + β x − 2 + χ x + 5 ⇔ x + 2 x (x − 2) (x + 5) = α (x − 2) (x + 5) + βx (x + 5) + χx (x − 2) x (x − 2) (x + 5) ⇔ x + 2 x (x − 2) (x + 5) = x 2 (α + β + χ) + x (3α + 5β − 2χ) − 10α x (x − 2) (x + 5) α + β + χ = 0 3α + 5β − 2χ = 1 −10α = 2 ⇔ α = − 1 5 β = 2 7 χ = − 3 35 x + 2 x (x − 2) (x + 5) = − 1 5x + 2 7 (x − 2) − 3 35 (x + 5) x + 2 x (x − 2) (x + 5) dx = − 1 5 dx x + 2 7 dx x − 2 − 3 35 dx x + 5 = − 1 5 ln |x| + 2 7 ln |x − 2| − 3 35 ln |x + 5| + c A 3 = x 2 (−3x 2 − 2x + 5) (x + 1) dx ax 2 + bx + c x 1 , , x 2 ax 2 + bx + c = a (x − x 1 ) (x − x 2 ) −3x 2 − 2x + 5 = −3 (x − 1) x + 5 3 x 2 (−3x 2 − 2x + 5) (x + 1) = − x 2 3 (x − 1) x + 5 3 (x + 1) = α x − 1 + β x + 5 3 + χ x + 1 α = lim x→1 − x 2 (x − 1) 3 (x − 1) x + 5 3 (x + 1) = lim x→1 − x 2 3 x + 5 3 (x + 1) = − 1 16 β = lim x→− 5 3 − x 2 x + 5 3 3 (x − 1) x + 5 3 (x + 1) = lim x→− 5 3 − x 2 3 (x − 1) (x + 1) = − 25 48 χ = lim x→−1 − x 2 (x + 1) 3 (x − 1) x + 5 3 (x + 1) = lim x→−1 − x 2 3 (x − 1) x + 5 3 = 1 4 x = 0 x 2 (−3x 2 − 2x + 5) (x + 1) = α x − 1 + β x + 5 3 + χ x + 1 ⇔ −α + 3 5 β + χ = 0 x = 2 x 2 (−3x 2 − 2x + 5) (x + 1) = α x − 1 + β x + 5 3 + χ x + 1 ⇔ − 4 33 = α + 3 11 β + 1 3 χ x = 3 x 2 (−3x 2 − 2x + 5) (x + 1) = α x − 1 + β x + 5 3 + χ x + 1 ⇔ − 9 112 = 1 2 α + 3 14 β + 1 4 χ −α + 3 5 β + χ = 0 α + 3 11 β + 1 3 χ = − 4 33 1 2 α + 3 14 β + 1 4 χ = − 9 112 ⇔ α = − 1 16 β = − 25 48 χ = 1 4 x 2 (−3x 2 − 2x + 5) (x + 1) = − x 2 3 (x − 1) x + 5 3 (x + 1) = α x − 1 + β x + 5 3 + χ x + 1 ⇔ − x 2 3 (x − 1) x + 5 3 (x + 1) = α x + 5 3 (x + 1) + β (x − 1) (x + 1) + χ (x − 1) x + 5 3 (x − 1) x + 5 3 (x + 1) ⇔ x 2 = x 2 (α + β + χ) + x 8 3 α + 2 3 χ + 5 3 α − β − 5 3 χ α + β + χ = 1 8 3 α + 2 3 χ = 0 5 3 α − β − 5 3 χ = 0 ⇔ α = − 1 16 β = − 25 48 χ = 1 4 x 2 (−3x 2 − 2x + 5) (x + 1) = − x 2 3 (x − 1) x + 5 3 (x + 1) = − 1 16 (x − 1) − 25 48 x + 5 3 + 1 4 (x + 1) x 2 (−3x 2 − 2x + 5) (x + 1) dx = − 1 16 dx x − 1 − 25 48 dx x + 5 3 + 1 4 dx x + 4 = − 1 16 ln |x − 1| − 25 48 ln x + 5 3 + 1 4 ln |x + 4| + c A 4 = x − 1 (x 2 + 4x + 5) (x 2 − 4) dx ax 2 + bx + c = 0 ∆ = b 2 − 4ac < 0 ax 2 + bx + c = a (x − x 1 ) (x − x 2 ) x 1 = α + βi, x 2 = α − βi, i 2 = −1 x 2 + 4x + 5 = (x + 2 + i) (x + 2 − i) x − 1 (x 2 + 4x + 5) (x 2 − 4) = x − 1 (x + 2 + i) (x + 2 − i) (x 2 − 4) = α x + 2 + i + β x + 2 − i + χ x − 2 + δ x + 2 α = lim x→−2−i (x − 1) (x + 2 + i) (x + 2 + i) (x + 2 − i) (x 2 − 4) = lim x→−2−i x − 1 (x + 2 − i) (x 2 − 4) = − 13 34 − 1 34 i β = lim x→−2+i (x − 1) (x + 2 − i) (x + 2 + i) (x + 2 − i) (x 2 − 4) = lim x→−2+i x − 1 (x + 2 + i) (x 2 − 4) = − 13 34 + 1 34 i χ = lim x→2 (x − 1) (x − 2) (x 2 + 4x + 5) (x − 2) (x + 2) = lim x→2 x − 1 (x 2 + 4x + 5) (x + 2) = 1 68 δ = lim x→−2 (x − 1) (x + 2) (x 2 + 4x + 5) (x − 2) (x + 2) = lim x→−2 x − 1 (x 2 + 4x + 5) (x − 2) = 3 4 x − 1 (x 2 + 4x + 5) (x 2 − 4) = − 13 34 − 1 34 i x + 2 + i + − 13 34 + 1 34 i x + 2 − i + 1 68 x − 2 + 3 4 x + 2 = −13x − 27 17 (x 2 + 4x + 5) + 1 68 (x − 2) + 3 4 (x + 2) x − 1 (x 2 + 4x + 5) (x 2 − 4) = αx + β x 2 + 4x + 5 + χ x − 2 + δ x + 2 x = 0 1 5 β − 1 2 χ + 1 2 δ = 1 20 x = 1 1 10 α + 1 10 β − χ + 1 3 δ = 0 x = 3 3 26 α + 1 26 β + χ + 1 5 δ = 1 65 x = 4 4 37 α + 1 37 β + 1 2 χ + 1 6 δ = 1 148 1 5 β − 1 2 χ + 1 2 δ = 1 20 1 10 α + 1 10 β − χ + 1 3 δ = 0 3 26 α + 1 26 β + χ + 1 5 δ = 1 65 4 37 α + 1 37 β + 1 2 χ + 1 6 δ = 1 148 ⇔ δ = − 2 5 β + χ + 1 10 1 10 α − 1 30 β − 2 3 χ = − 1 30 3 26 α − 27 650 β + 6 5 χ = − 3 650 4 37 α − 22 555 β + 2 3 χ = − 11 1110 ⇔ α = − 13 17 β = − 27 17 χ = 1 68 δ = 3 4 x − 1 (x 2 + 4x + 5) (x 2 − 4) = −13x − 27β 17 (x 2 + 4x + 5) + χ 68 (x − 2) + 3δ 4 (x + 2) x − 1 (x 2 + 4x + 5) (x 2 − 4) = αx + β x 2 + 4x + 5 + χ x − 2 + δ x + 2 ⇔ x − 1 = (αx + β) x 2 − 4 + χ x 2 + 4x + 5 (x + 2) + δ x 2 + 4x + 5 (x − 2) ⇔ x − 1 = x 3 (α + χ + δ) + x 2 (β + 6χ + 2δ) + x (−4α + 13χ − 3δ) + (−4β + 10χ − 10δ) α + χ + δ = 0 β + 6χ + 2δ = 0 −4α + 13χ − 3δ = 1 4β − 10χ + 10δ = 1 ⇔ δ = −α −χ β + 6χ + 2 (−α − χ) = 0 −4α + 13χ − 3 (−α − χ) = 1 4β − 10χ + 10 (−α − χ) = 1 ⇔ δ = −α −χ −2α + β + 4χ = 0 −α + 16χ = 1 −10α + 4β − 20χ = 1 ⇔ α = − 13 17 β = − 27 17 χ = 1 68 δ = 3 4 x − 1 (x 2 + 4x + 5) (x 2 − 4) = −13x − 27 17 (x 2 + 4x + 5) + 1 68 (x − 2) + 3 4 (x + 2) x − 1 (x 2 + 4x + 5) (x 2 − 4) dx = − 13 34 2x + 54 13 x 2 + 4x + 5 dx + 1 68 dx x − 2 + 3 4 dx x + 2 = − 13 34 (2x + 4) x 2 + 4x + 5 dx − 1 17 dx x 2 + 4x + 5 + 1 68 dx x − 2 + 3 4 dx x + 2 = − 13 34 d x 2 + 4x + 5 x 2 + 4x + 5 − 1 17 dx (x + 2) 2 + 1 + 1 68 dx x − 2 + 3 4 dx x + 2 = − 13 34 ln x 2 + 4x + 5 − 1 17 arctan (x + 2) + 1 68 ln |x − 2| + 3 4 ln |x + 2| + c A 5 = x x 3 + 1 dx x x 3 + 1 = x (x + 1) (x 2 − x + 1) = α x + 1 + β x − 1 2 − √ 3 2 i + χ x − 1 2 + √ 3 2 i α = lim x→−1 x x 2 − x + 1 = − 1 3 β = lim x→ 1 2 + √ 3 2 i x (x + 1) x − 1 2 + √ 3 2 i = 1 6 − √ 3 6 i χ = lim x→ 1 2 − √ 3 2 i x (x + 1) x − 1 2 − √ 3 2 i = 1 6 + √ 3 6 i x x 3 + 1 = − 1 3 (x + 1) + 1 6 − √ 3 6 i x − 1 2 − √ 3 2 i + 1 6 + √ 3 6 i x − 1 2 + √ 3 2 i = − 1 3 (x + 1) + x + 1 3 (x 2 − x + 1) x x 3 + 1 = α x + 1 + βx + χ x 2 − x + 1 x = 0 0 = α + χ x = 1 1 2 = 1 2 α + β + χ x = 2 2 9 = 1 3 α + 2 3 β + 1 3 χ α + χ = 0 1 2 α + β + χ = 1 2 1 3 α + 2 3 β + 1 3 χ = 2 9 ⇔ α = − 1 3 β = 1 3 χ = 1 3 x x 3 + 1 = − 1 3 (x + 1) + x + 1 3 (x 2 − x + 1) x x 3 + 1 = α x + 1 + βx + χ x 2 − x + 1 ⇔ x = α x 2 − x + 1 + (βx + χ) (x + 1) α + β = 0 −α + β + χ = 1 α + χ = 0 ⇔ α = − 1 3 β = 1 3 χ = 1 3 x x 3 + 1 dx = − 1 3 dx x + 1 + 1 6 2x + 2 x 2 − x + 1 dx = − 1 3 dx x + 1 + 1 6 2x + 1 x 2 − x + 1 dx + 1 6 dx x 2 − x + 1 = − 1 3 dx x + 1 + 1 6 d x 2 − x + 1 x 2 − x + 1 + 1 6 dx x − 1 2 2 + 3 4 = − 1 3 ln |x + 1| + 1 6 ln x 2 − x + 1 + 1 3 √ 3 arctan 2x − 1 √ 3 + c A 6 = dx x 8 + 1 x 8 + 1 = x 4 + px 2 + 1 x 4 − px 2 + 1 = x 8 + 2 − p 2 x 4 + 1 2 − p 2 = 0 ⇔ p = ± √ 2 ⇒ x 8 + 1 = x 4 + √ 2x 2 + 1 x 4 − √ 2x 2 + 1 x 4 + √ 2x 2 + 1 = x 2 + qx + 1 x 2 − qx + 1 = x 4 + 2 − q 2 x 2 + 1 2 − q 2 = √ 2 ⇔ q = ± 2 − √ 2 ⇒ x 4 + √ 2x 2 + 1 = x 2 + 2 − √ 2x + 1 x 2 − 2 − √ 2x + 1 x 4 − √ 2x 2 + 1 = x 2 + rx + 1 x 2 − rx + 1 = x 4 + 2 − r 2 x 2 + 1 2 − r 2 = − √ 2 ⇔ r = ± 2 + √ 2 ⇒ x 4 − √ 2x 2 + 1 = x 2 + 2 + √ 2x + 1 x 2 − 2 + √ 2x + 1 x 8 + 1 = x 2 + 2 − √ 2x + 1 x 2 − 2 − √ 2x + 1 x 2 + 2 + √ 2x + 1 x 2 − 2 + √ 2x + 1 ⇒ 1 x 8 + 1 = 1 8 2 − √ 2x + 2 x 2 + 2 − √ 2x + 1 + 1 8 − 2 − √ 2x + 2 x 2 − 2 − √ 2x + 1 + 1 8 2 + √ 2x + 2 x 2 + 2 + √ 2x + 1 + + 1 8 − 2 + √ 2x + 2 x 2 − 2 + √ 2x + 1 dx ax 2 + bx + c = 1 a dx x 2 + b a x + c a , 4ac − b 2 > 0 = 1 a dx x − −b 2a 2 + 4ac − b 2 2a 2 = 1 a dx (x − p) 2 + q 2 (1) p = −b 2a , q = √ 4ac − b 2 2a , x = p + qt = 1 aq 1 t 2 + 1 dt = 2 √ 4ac − b 2 1 t 2 + 1 dt = 2 √ 4ac − b 2 arctan t = 2 √ 4ac − b 2 arctan 2ax + b √ 4ac − b 2 + C t = x − p q = 2ax + b √ 4ac − b 2 xdx ax 2 + bx + c = 1 2a 2ax + b − b ax 2 + bx + c dx = 1 2a 2ax + b ax 2 + bx + c dx − b 2a 1 ax 2 + bx + c dx = 1 2a ln ax 2 + bx + c − b 2a dx ax 2 + bx + c + C a = c = 1, ∆ = 4 − b 2 > 0 Ax + B x 2 + bx + 1 dx = A xdx x 2 + bx + 1 + B dx x 2 + bx + 1 = 2B −Ab √ 4 − b 2 arctan 2x + b √ 4 − b 2 + A 2 ln x 2 + bx + 1 + C A 6.1 = 1 8 2 − √ 2x + 2 x 2 + 2 − √ 2x + 1 dx = = 2 + √ 2 8 arctan 2x + 2 − √ 2 2 + √ 2 + 2 − √ 2 16 ln x 2 + 2 − √ 2x + 1 + C 1 A 6.2 = 1 8 − 2 − √ 2x + 2 x 2 − 2 − √ 2x + 1 dx = = 2 + √ 2 8 arctan 2x − 2 − √ 2 2 + √ 2 − 2 − √ 2 16 ln x 2 − 2 − √ 2x + 1 + C 2 A 6.3 = 2 + √ 2x + 2 x 2 + 2 + √ 2x + 1 dx = = 2 − √ 2 8 arctan 2x + 2 − √ 2 2 − √ 2 + 2 + √ 2 16 ln x 2 + 2 + √ 2x + 1 + C 3 A 6.4 = − 2 + √ 2x + 2 x 2 − 2 + √ 2x + 1 dx = = 2 − √ 2 8 arctan 2x − 2 − √ 2 2 − √ 2 − 2 + √ 2 16 ln x 2 − 2 + √ 2x + 1 + C 4 A 6 = A 6.1 + A 6.2 + A 6.3 + A 6.4 x 8 + 1 = 0 ⇔ x 8 = −1 = cos (π + k2π) + i sin (π + k2π) ⇒ x = cos π + k2π 8 + i sin π + k2π 8 k = 0, , 7 sin π 8 = sin 7π 8 = cos 3π 8 = 1 2 2 − √ 2, sin 3π 8 = sin 5π 8 = cos π 8 = 1 2 2 + √ 2 , cos 5π 8 = cos 7π 8 = − 1 2 2 − √ 2 , dx 1 + x 8 = − 1 8 3 k=0 ln x 2 − 2x cos (2k + 1) π 8 + 1 × cos (2k + 1) π 8 + + 1 4 3 k=0 arctan x sin (2k + 1) π 8 1 − x cos (2k + 1) π 8 × sin (2k + 1) π 8 + C, k = 0, , 7