Sol_Less10 - 10 ‫תרגיל‬ ‫שרמקוב...

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Unformatted text preview: 10 ‫תרגיל‬ ‫שרמקוב יוליה‬ 3 1 2 x2 23 xdx = ∫ x dx = + c = x2 + c 3 3 2 ∫ n n m n +1 .1 +1 xm mx m mn x dx = ∫ x dx = +c = + c .2 ∫ n n+m +1 m dx x −1 1 = ∫ x −2 dx = + c = − + c .3 ∫ x2 −1 x 1 dx 1 −1 1 x2 = ∫ x 2 dx = + c = x + c .4 ∫2 x 2 21 2 −0.17 −0.17 ∫ 3.4 x dx = 3.4∫ x dx = 3.4 ×(−0.17) x −0.17 +1 + c ≈ 4.1x 0.83 + c .5 −0.17 + 1 dx 1 ∫ 2x − 3 = 2 ∫ ∫( )( dx 1 3 = ln x − ÷+ c 32 .6 2 x− 2 ) x + 1 x − x + 1 dx = ∫ ( x × + x − x − x + x + 1)dx = x 3 = ∫ ( x × + 1)dx = ∫ x 2 dx + ∫ dx = x 52 x x +x+c 2 ( ae ) ( ae ) ( ae ) ∫ a e dx = ∫ ( ae ) dx = ln ( ae ) + c = ln e + ln a + c = 1 + ln a + c x x x x xx .7 .8 x 3 2 ÷ x x x x x 3 ×2 − 2 × 3 3 ×2 2× 3 3 2 ∫ 2 x dx = ∫ 2 x dx − ∫ 2 x dx = 3∫ dx − 2∫ 2 ÷ dx = 3x − 3 + c .9 ln 2 ∫ dx 3 − 3x 2 1 = 1 dx 1 ∫ 1 − x 2 = 3 arcsin x + c 3 .10 10 ‫תרגיל‬ ‫שרמקוב יוליה‬ ( 1+ x) 2 ∫ x ( 1+ x ( 1 + 2 x + x ) dx = ( 1 + x ) dx + 2 xdx = =∫ ∫ x (1+ x ) ∫ x (1+ x ) x ( 1+ x ) ) dx 2 2 2 2 2 2 dx dx = ∫ + 2∫ = ln x + 2 arctan x + c x 1 + x2 ( .11 ) 1 + cos 2 x 1 + cos 2 x 1 + cos 2 x 1 1 + cos 2 x dx = ∫ dx = ∫ dx = ∫ dx = ∫ 1 + cos 2 x 1 + cos 2 x − sin 2 x 2 cos 2 x 2 cos 2 x .12 1 1 1 1 1 =∫ dx + ∫ dx = tan x + x + c 2 cos 2 x 2 2 2 cos 2 x cos 2 x − sin 2 x cos 2 x sin 2 x dx = ∫ dx = ∫ dx − ∫ dx = ∫ cos2 x ×sin 2 x cos 2 x × 2 x sin cos 2 x × 2 x sin cos 2 x × 2 x sin .13 1 1 = ∫ 2 dx − ∫ dx = − cot x − tan x + c sin x cos 2 x ∫ tan 2 1 1 xdx = ∫ − 1÷ = ∫ dx dx ÷ − ∫ dx = tan x − x + c .14 2 2 cos x cos x ∫ 2sin 2 x dx = ∫ (1 − cos x) dx = ∫ dx − ∫ cos xdx = x − sin x + c .15 2 dx ∫ cos 2 x + sin 2 x =∫ dx dx =∫ = tan x + c .16 2 2 cos x − sin x + sin x cos 2 x 2 { ∫ sin xd ( sin x ) = t = sin x, dt = d ( sin x ) } = ∫ tdt = { 3 3 ∫ tan xd ( tan x ) = t = tan x, dt = d ( tan x ) } = ∫ t dt = t4 1 + c = tan 4 x + c .18 4 4 3 3 ∫ sin x cos xdx = { t = sin x, dt = cos xdx} = ∫ t dt = ∫ cos 3 t2 1 + c = sin 2 x + c .17 2 2 t4 1 + c = sin 4 x + c .19 4 4 x sin 2 xdx = ∫ cos 3 x(2sin x cos x) dx = 2 ∫ cos 4 x sin xdx = = { t = cos x, dt = − sin xdx} = −2 ∫ t 4 dt = − 2 .20 2t 5 2 + c = − cos5 x + c 5 5 10 ‫תרגיל‬ ‫שרמקוב יוליה‬ ∫ cos x 3 sin 2 x dx = { t = sin x, dt = cos xdx} = ∫ dt 3 t2 =∫ dt t 2 3 1 3 1 3 = 3t + c = 3sin x + c .21 1 1 ∫ sin ( 2 x − 3) dx = t = 2 x − 3 ⇒ dt = 2dx ⇒ 2 dt = dx = 2 ∫ sin tdt = 1 1 = − cos t + c = − cos ( 2 x − 3) + c 2 2 sin x − cos x ∫ sin x + cos x dx = { t = sin x + cos x ⇒ −dt = (sin x − cos x)dx} = − ∫ .22 dt = t .23 = − ln t + c = − ln sin x + cos x + c ∫ ( d 1 + x2 1+ x ∫ 2 ) = { t =1 + x , dt = d 2 ( 1 + x ) } = ∫ dtt = 2 2 t + c = 2 1 + x 2 + c .24 1 11 8 − 2 xdx = { t = 8 − 2 x, dt = −2dx ⇒ − dt = dx = − ∫ t 2 dt = 2 2 3 2 3 2 1 2t (8 − 2 x) = − × +c = − +c 23 3 .25 x3 1 1 1 4 3 3 ∫ x8 + 1dx = { t = x , dt = 4 x dx ⇒ 4 dt = x dx = 4 ∫ t 2 + 1dt = .26 1 1 4 = arctan t + c = arctan x + c 4 4 ∫ 5 1 1 (8 − 3x )6 dx = { t = 8 − 3 x ⇒ dt = −3dx ⇒ − dt = dx = − ∫ 5 t 6 dx = 3 3 11 5 11 5 = − t 5 + c = − (8 − 3 x) 5 + c 33 33 x 2 + 1dx = { t = x 2 + 1 ⇒ dt = 2 xdx} = ∫ tdt = ∫ 2x ∫ x 3 3 x +1 4 dx = { t = x 4 + 1 ⇒ dt = 4 x 3dx 1 dt }= ∫3 4 t 3 23 2 t +c = 3 3 = 2 3 3t +c = 8 (x 2 ( ) 3 + 1 + c .28 ) 3 x4 + 1 8 .27 2 3 + c .29 10 ‫תרגיל‬ ‫שרמקוב יוליה‬ ∫ 3 ln x 1 23 2 dx = { t = ln x ⇒ dt = dx = ∫ tdt = t 2 + c = ( ln x ) 2 + c x x 3 3 .30 e x dx dt x x x ∫ e x + 1 = { t = e + 1 ⇒ dt = e dx = ∫ t = ln t + c = ln e + 1 + c .31 } ( ) e 2 x dx 1 dt 1 1 2x 2 2x 2x 2 ∫ e2 x + a 2 = { t = e + a ⇒ dt = 2e dx = 2 ∫ t = 2 ln t + c = 2 ln e + a + c .32 } ( ) arctan 2 xdx dx t3 arctan 3 x 2 ∫ 1 + x 2 = { t = arctan x ⇒ dt = 1 + x 2 = ∫ t dt = 3 + c = 3 + c x 2 dx 1 dt 1 3 2 2 ∫ 4 + x6 = { t = x ⇒ dt = 3x dx ⇒ 3 dt = x dx = 3 ∫ 4 + t 2 = 1 t 1 x3 = arctan + c = arctan + c 6 2 6 2 xdx ∫ 4+ x ={t =x 4 2 ⇒ dt = 2 xdx ⇒ 1 dt 1 dt = xdx = ∫ = 2 2 2 4+t 1 t 1 x2 = arctan + c = arctan + c 4 2 4 2 .33 .34 .35 dx 1 dx 1 dx 2 2x =∫ =∫ = arctan +c 2 2 + 9 2 x2 + 9 2 2 3 6 3 .36 x + ÷ 2 2 ∫ 2x ∫ ∫ ∫ dy 6 y2 + 2 dx 3x 2 + 7 = = 1 6∫ 1 3∫ 4 dy y2 + 1 3 dx x2 + 7 3 dt t +4 2 ( ) = ln t + t 2 + 4 + c .37 = 1 1 ln y + y 2 + ÷+ c .38 3÷ 6 = 1 7 ln x + x 2 + ÷+ c .39 3÷ 3 10 ‫תרגיל‬ ‫שרמקוב יוליה‬ dz ∫ ∫ dx 4− x 2 = 4z 2 − 9 1 2∫ dx 2 x 1− ÷ 2 x = arcsin t + c = arcsin + c 2 dx ∫ 4 − 9x2 = ∫ = 1 2∫ 1 2∫ ={t = dx 2 3x 1− ÷ 2 = dz z2 − 9 4 = 1 9 ln z + z 2 − ÷+ c .40 2 4÷ x 1 ⇒ dt = dx 2 2 ={t = } =∫ dt 1− ( t ) 2 = .41 3x 3 2 ⇒ dt = dx ⇒ dt = dx = 2 2 3 .42 1 dt 1 1 3x = arcsin t + c = arcsin + c ∫ 2 3 1− ( t ) 3 3 2 2 x dx 1− 4 x =∫ 2 x dx () 1− 2 x 2 = { t = 2 x ⇒ dt = 2 x ln 2dx ⇒ dt = 2 x dx ln 2 }= .43 1 dt 1 1 = = arcsin t + c = arcsin 2 x + c ∫ ln 2 1 − ( t ) 2 ln 2 ln 2 dx ∫ ( arcsin x ) 2 1− x 2 = { t = arcsin x ⇒ dt = = 1− x dx 2 t −1 1 = ∫ t −2 dt = +c = − +c −1 arcsin x .44 :‫שיטת אינטגרציה לפי חלקים‬ 1 u = ln ( x − 1) du = dx x ln ( x − 1) dx = ⇒ dx = x − 1 = x ln ( x − 1) − ∫ ∫ x −1 dv = dx v=x x −1+1 x −1 1 dx =x ln ( x − 1) − ∫ dx − ∫ dx = x −1 x −1 x −1 1 = x ln ( x − 1) − ∫ dx − ∫ dx = x ln ( x − 1) − x − ln ( x − 1) + c x −1 = x ln ( x − 1) − ∫ 5 .45 10 ‫תרגיל‬ ‫שרמקוב יוליה‬ 1 dx 3 u = ln x 1 x3 xx 2 x ln xdx = ⇒ = ln ( x ) − ∫ dx = 2 ∫ 1 3 3x dv = x dx v = x3 .46 3 du = = x3 1 x3 1 ln ( x ) − ∫ x 2 dx = ln ( x ) − x 3 + c 3 3 3 9 2x du = 2 dx u = ln x 2 + 1 2 x +1 = ∫ ln x + 1 dx = dv = dx ⇒ v=x 2 2 x x +1−1 = x ln x 2 + 1 − 2 ∫ 2 dx = x ln x 2 + 1 − 2∫ 2 dx = .47 x +1 x +1 ( ( ) ( ) ( ) ( ( ) = x ln x 2 + 1 − 2 ∫ ( ) ( ( ) ) ) ( ) x2 + 1 1 dx + 2 ∫ 2 dx = 2 x +1 x +1 ) ( ) = x ln x 2 + 1 − 2 x + 2 arctan x + c 2 dx u = arctan 2 x du = 2 4x +1 = ∫ arctan 2 xdx = dv = dx ⇒ v=x .48 2x 1 = x arctan 2 x − ∫ dx = x arctan 2 x − ln 4 x 2 + 1 + c 4 4x2 + 1 ( ( ) ( ) u = arctan x x ×arctan xdx = ⇒ ∫ dv = xdx 1 dx x +1 = 1 v = x2 2 du = ( 2 ) ) ( ) 2 12 1 x2 12 1 x +1 −1 = x arctan x − ∫ 2 dx = x arctan x − ∫ dx = 2 2 x +1 2 2 x2 + 1 ( = ) ( ) 12 1 1 1 1 1 1 x arctan x − x + ∫ 2 dx = x 2 arctan x − x + arctan x + c 2 2 2 x +1 2 2 2 ( ) 6 .49 10 ‫תרגיל‬ ‫שרמקוב יוליה‬ du = 3 x 2 dx u = x3 1 3 5x 3 2 5x ∫ x ×e dx = dv = e5 x dx ⇒ v = 1 e5 x = 5 x ×e − 5 ∫ x ×e dx = 5 du1 = 2 xdx u1 = x 2 1 3 5x 3 1 2 5x 2 5x = ⇒ 1 5 x = x ×e − x ×e − ∫ x ×e dx = 5x 5 5 5 v= e 5 dv = e dx 5 .50 du2 = dx u2 = x 1 3 5x 3 2 5x 6 = x ×e − x ×e + ∫ x ×e5 x dx = ⇒ 1 5x = 5x 5 25 25 dv = e dx v = e 5 1 3 6 1 1 = x 3 ×e5 x − x 2 ×e5 x + e5 x ×x − ∫ e5 x dx = 5 25 25 5 5 1 3 6 1 5x = x 3 ×e5 x − x 2 ×e5 x + x ×e5 x − e +c 5 25 125 625 3 5x u = x3 du = 3 x 2 dx 3 −x 2 −x ∫ x ×e dx = dv = e− x dx ⇒ v = −e− x = − x ×e + 3∫ x ×e dx = 2 du = 2 xdx u =x = 1 −x ⇒ 1 = − x 3 ×e − x + 3 − x 2 ×e − x + 2 ∫ x ×e − x dx = −x v = −e dv = e dx −x 3 du = dx u =x = − x ×e − 3 x ×e + 6 ∫ x ×e dx = 2 − x ⇒ 2 − x = dv = e dx v = −e = − x 3 ×e − x − 3 x 2 ×e − x + 6 −e − x ×x + ∫ e − x dx = 3 −x 2 −x −x −x = − x ×e − 3 x ×e − 6 x ×e − 6e + c 3 −x 2 −x .51 −x u=x du = dx ∫ x ×cos xdx = dv = cos xdx ⇒ v = sin x = x ×sin x − ∫ sin xdx = .52 = x × x + cos x + c sin du = 2 xdx u = x2 2 ∫ x ×sin xdx = dv = sin xdx ⇒ v = − cos x = − x ×cos x + 2∫ x cos xdx = .53 du = dx u1 = x 2 = ⇒1 = − x ×cos x + 2 x sin x − ∫ sin xdx = dv1 = cos xdx v1 = sin x 2 = − x 2 ×cos x + 2 x sin x + 2 cos x + c 7 10 ‫תרגיל‬ ‫שרמקוב יוליה‬ u=x du = dx x ∫ cos2 x dx = dv = dx ⇒ v = tan x = x ×tan x − ∫ tan xdx = cos 2 x .54 sin x 1 = x ×tan x − ∫ dx = { t = cos x ⇒ dt = − sin xdx} = x ×tan x + ∫ dt = cos x t = x ×tan x + ln t + c = x ×tan x + ln cos x + c .55 du = cos xdx u = sin x 1 2x 1 2x ∫ e sin xdx = dv = e2 x dx ⇒ v = 1 e2 x = 2 e sin x − 2 ∫ e cos xdx = 2 du1 = − sin xdx u1 = cos x 1 1 2x 1 2x 1 2x = ⇒ 1 2 x = e sin x − e cos x + ∫ e sin xdx = 2x 2 2 2 v= e dv = e dx 2 2 1 1 1 = e 2 x sin x − e 2 x cos x − ∫ e 2 x sin xdx ⇒ A = ∫ e 2 x sin xdx } 2 4 4 1 1 1 ⇒ A = e 2 x sin x − e 2 x cos x − A 2 4 4 5 1 1 ⇒ A = e 2 x sin x − e 2 x cos x + c 4 2 4 2 1 ⇒ A = e 2 x sin x − e 2 x cos x + c 5 5 2 1 ⇒ ∫ e 2 x sin xdx = e 2 x sin x − e 2 x cos x + c 5 5 2x { 8 10 ‫תרגיל‬ ‫שרמקוב יוליה‬ tdt u = t 2 + 4 du = t 2 dt 2 t + 4dt = ⇒ = t t2 + 4 − ∫ = t +4 t2 + 4 dv = dt v=t ∫ 2 =t t +4−∫ 2 (t 2 ) + 4 − 4 dt =t t +4−∫ 2 t +4 2 = t t 2 + 4 − ∫ t 2 + 4dt + 4 ∫ dt t +4 2 (t 2 ) + 4 dt t +4 2 + 4∫ dt t +4 2 = t t2 + 4 − − ∫ t + 4dt + 4 ln(t + t + 4) = 2 2 { = A = ∫ t 2 + 4dt = .56 } ⇒ A = t t 2 + 4 − A + 4 ln(t + t 2 + 4) ⇒ 2 A = t t 2 + 4 + 4 ln(t + t 2 + 4) + c 1 ⇒ A = t t 2 + 4 + 2 ln(t + t 2 + 4) + c 2 1 ⇒ ∫ t 2 + 4dt = t t 2 + 4 + 2 ln(t + t 2 + 4) + c 2 .57 xdx u = x 2 − 2 du = x 2 dx 2 x − 2dx = ⇒ = x x2 − 2 − ∫ = x −2 dv = dx x2 − 2 v=x ∫ 2 = x x −2 −∫ 2 (x 2 ) − 2 + 2 dx x2 − 2 = x x 2 − 2 − ∫ x 2 − 2dx − 2 ∫ { = A = ∫ x 2 − 2dx = x x −2 −∫ 2 dx x −2 2 (x 2 ) − 2 dx x2 − 2 − 2∫ dx x2 − 2 = = x x 2 − 2 − ∫ x 2 − 2dx − 2 ln( x + x 2 − 2) = } ⇒ A = x x 2 − 2 − A − 2 ln( x + x 2 − 2) ⇒ 2 A = x x 2 − 2 − 2 ln( x + x 2 − 2) + c 1 ⇒ A = x x 2 − 2 − ln( x + x 2 − 2) + c 2 1 ⇒ ∫ x 2 − 2dx = x x 2 − 2 − ln( x + x 2 − 2) + c 2 9 ...
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