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Question Number 91099 by Angnysr11 last updated on 28/Apr/20

Commented by Angnysr11 last updated on 28/Apr/20

i determinant (( [((i can′t solve this... please help)),() ]),())

i|[icantsolvethis...pleasehelp]|

Commented by john santu last updated on 28/Apr/20

15) ∫_4 ^8  5u(√(4+2u)) du =  (√(4+2u)) = w ⇒ 2du = 2w dw   ∫^(2(√5)) _(2(√3)) 5(((w^2 −4)/2)).w wdw =   (5/2)∫_(2(√3)) ^(2(√5)) (w^4 −4w^2 )dw =  (5/2)[(1/5)w^5 −(4/3)w^3  ]_(2(√3)) ^(2(√5))   now it easy  to solve

15)845u4+2udu=4+2u=w2du=2wdw23255(w242).wwdw=522325(w44w2)dw=52[15w543w3]2325nowiteasytosolve

Commented by abdomathmax last updated on 28/Apr/20

2) ∫ 10x(5x+3)^3  dx =10 .5^3  ∫x(x+(3/5))^3  dx  =_(by parts)   10.5^3 {  (x/4)(x+(3/5))^4 −∫  (1/4)(x+(3/5))^4  dx}  =20.5^3 { (x/4)(x+(3/5))^4 −(1/(20)) (x+(3/5))^5 } +C

2)10x(5x+3)3dx=10.53x(x+35)3dx=byparts10.53{x4(x+35)414(x+35)4dx}=20.53{x4(x+35)4120(x+35)5}+C

Commented by abdomathmax last updated on 28/Apr/20

1) by parts  ∫ x(x−2)^4  dx =((x(x−2)^5 )/5)  −∫(((x−2)^5 )/5)dx =((x(x−2)^5 )/5)−(1/(30))(x−2)^6  +C

1)bypartsx(x2)4dx=x(x2)55(x2)55dx=x(x2)55130(x2)6+C

Commented by abdomathmax last updated on 28/Apr/20

3) ∫_2 ^5 x(√(x+3))dx  =_((√(x+3))=t)   ∫_(√5) ^(2(√2)) (t^2 −3)t (2t)dt  =2 ∫_(√5) ^(2(√2)) (t^4 −3t^2 )dt =2[(t^5 /5)−t^3 ]_(√5) ^(2(√2))   =2((((2(√2))^5 )/5)−(2(√2))^3 −((((√5))^5 )/5) +((√5))^3 )

3)25xx+3dx=x+3=t522(t23)t(2t)dt=2522(t43t2)dt=2[t55t3]522=2((22)55(22)3(5)55+(5)3)

Commented by abdomathmax last updated on 28/Apr/20

4) ∫_1 ^3  3x^2 (√(x+7))dx =_(.(√(x+7))=t)   ∫_(2(√2)) ^(√(10)) 3(t^2 −7)^2 t(2t)dt  =6 ∫_(2(√2)) ^(√(10)) t^2 (t^4 −14t^2 +49)dt  =6 ∫_(2(√2)) ^(√(10)) (t^6 −14t^4  +49t^2 )dt  =6[(t^7 /7)−((14)/5)t^5  +((49)/3)t^3 ]_(2(√2)) ^(√(10))   =6( ((((√(10)))^7 )/7)−((14)/5)((√(10)))^5  +((49)/3)((√(10)))^3 −(((2(√2))^7 )/7)+((14)/5)(2(√2))^5   −((49)/3)(2(√2))^3 )

4)133x2x+7dx=.x+7=t22103(t27)2t(2t)dt=62210t2(t414t2+49)dt=62210(t614t4+49t2)dt=6[t77145t5+493t3]2210=6((10)77145(10)5+493(10)3(22)77+145(22)5493(22)3)

Commented by abdomathmax last updated on 28/Apr/20

10) ∫ x^2 (^3 (√(x+1)))dx  we do the changement^3 (√(x+1))=t  x+1 =t^3  ⇒ I =∫ (t^3 −1)^2 t (3t^2 )dt  =3 ∫ t^3 (t^6 −2t^3 +1)dt  =3∫ (t^9 −2t^6  +t^3 )dt =3((t^(10) /(10))−(2/7)t^7  +(t^4 /4)) +C  =(3/(10))t^(10)  −(6/7)t^7  +(3/4) t^4  +C  =(3/(10))(^3 (√(x+1)))^(10)  −(6/7)(^3 (√(x+1)))^7  +(3/4)(^3 (√(x+1)))^(4 )  +C

10)x2(3x+1)dxwedothechangement3x+1=tx+1=t3I=(t31)2t(3t2)dt=3t3(t62t3+1)dt=3(t92t6+t3)dt=3(t101027t7+t44)+C=310t1067t7+34t4+C=310(3x+1)1067(3x+1)7+34(3x+1)4+C

Commented by abdomathmax last updated on 28/Apr/20

11) ∫  (x/(√(x+2)))dx we do the changement (√(x+2))=t ⇒  ∫  (x/(√(x+2)))dx =∫  ((t^2 −2)/t)(2t)dt  =2 ∫(t^2 −2)dt =2((t^3 /3)−2t) +C  =(2/3)t^3 −4t +C  =(2/3)((√(x+2)))^3 −4(√(x+2)) +C

11)xx+2dxwedothechangementx+2=txx+2dx=t22t(2t)dt=2(t22)dt=2(t332t)+C=23t34t+C=23(x+2)34x+2+C

Commented by abdomathmax last updated on 28/Apr/20

12) ∫ x^2 (√(4−x))dx we do the changement (√(4−x))=u ⇒  4−x =u^2  ⇒x =4−u^2  ⇒  ∫ x^2 (√(4−x))dx =∫ (4−u^2 )^2 u (−2u)du  =−2∫ u^2 (u^4 +8u^2  +16)du  =−2∫ (u^6  +8u^4  +16u^2 )du  =−2( (u^7 /7) +(8/5)u^5  +((16)/3)u^2 ) +C  =−(2/7)((√(4−x)))^7 −((16)/5)((√(4−x)))^5  −((32)/3)((√(4−x)))^2  +C

12)x24xdxwedothechangement4x=u4x=u2x=4u2x24xdx=(4u2)2u(2u)du=2u2(u4+8u2+16)du=2(u6+8u4+16u2)du=2(u77+85u5+163u2)+C=27(4x)7165(4x)5323(4x)2+C

Commented by mathmax by abdo last updated on 28/Apr/20

13) ∫_1 ^4 x(x+5)^3 dx =_(by parts)   [(x/4)(x+5)^4 ]_1 ^4 −∫_1 ^4 (1/4)(x+5)^4  dx  =9^4 −(6^4 /4)−(1/(20))[(x+5)^5 ]_1 ^4  =9^4 −(6^4 /4)−(1/(20))( 9^5 −6^5 )

13)14x(x+5)3dx=byparts[x4(x+5)4]141414(x+5)4dx=94644120[(x+5)5]14=94644120(9565)

Commented by mathmax by abdo last updated on 28/Apr/20

18)  I =∫ (x+1)(x+5)^(−3)  dx  by parts u =x+1 and v^′ =(x+5)^(−3)   I =(1/(−2))(x+1)(x+5)^(−2)  +(1/2)∫  (x+5)^(−2)  dx  =−(((x+1))/(2(x+5)^2 )) +(1/2)∫  (dx/((x+5)^2 )) =−((x+1)/(2(x+5)^2 ))−(1/(2(x+5))) +C

18)I=(x+1)(x+5)3dxbypartsu=x+1andv=(x+5)3I=12(x+1)(x+5)2+12(x+5)2dx=(x+1)2(x+5)2+12dx(x+5)2=x+12(x+5)212(x+5)+C

Commented by mathmax by abdo last updated on 28/Apr/20

15) I =∫_4 ^8  5x(√(4+2x))dx  we do the changement (√(4+2x))=t ⇒  4+2x =t^2  ⇒2x =t^2 −4 ⇒x =(t^2 /2)−2 ⇒  I =5 ∫_(2(√3)) ^(2(√5))   ((t^2 /2)−2)t (t)dt =5 ∫_(2(√3)) ^(2(√5)) ((t^4 /2)−2t^2 )dt  =5[(1/(10))t^5 −(2/3)t^3 ]_(2(√3)) ^(2(√5))   =5(  (((2(√5))^5 )/(10))−(2/3)(2(√5))^3 −(((2(√3))^5 )/(10))+(2/3)(2(√3))^3 )

15)I=485x4+2xdxwedothechangement4+2x=t4+2x=t22x=t24x=t222I=52325(t222)t(t)dt=52325(t422t2)dt=5[110t523t3]2325=5((25)51023(25)3(23)510+23(23)3)

Answered by Kunal12588 last updated on 28/Apr/20

1. ∫u(u−2)^4 du=∫(u−2)^5 du+2∫(u−2)^4 du         = (1/6)(u−2)^6 +(2/5)(u−2)^5 +C  2. ∫10u(5u+3)^3 du=2∫(5u+3)^4 du−6∫(5u+3)^3 du          =(2/(25))(5u+3)^5 −(3/(10))(5u+3)^4 +C  3. ∫_2 ^5 u(√(u+3)) du=∫_2 ^5 (u+3)^(3/2) du−3∫_2 ^5 (u+3)^(1/2) du         =(2/5)[(u+3)^2 (√(u+3))]_2 ^5 −2[(u+3)(√(u+3))]_2 ^5          =(2/5)(128(√2)−25(√5))−2(16(√2)−5(√5))         =((96(√2))/5)  4. ∫_1 ^3 3u^2 (√(u+7)) du=∫_1 ^3 3((u+7)−7)^2 (√(u+7)) du  =3[∫_1 ^3 (u+7)^(5/2) du−14∫_1 ^3 (u+7)^(3/2) du+49∫(u+7)^(1/2) du]  =(6/7)[(u+7)^(7/2) ]_1 ^3 −((84)/5)[(u+7)^(5/2) ]_1 ^3 +49[(u+7)^(3/2) ]_1 ^3   =...

1.u(u2)4du=(u2)5du+2(u2)4du=16(u2)6+25(u2)5+C2.10u(5u+3)3du=2(5u+3)4du6(5u+3)3du=225(5u+3)5310(5u+3)4+C3.25uu+3du=25(u+3)3/2du325(u+3)1/2du=25[(u+3)2u+3]252[(u+3)u+3]25=25(1282255)2(16255)=96254.133u2u+7du=133((u+7)7)2u+7du=3[13(u+7)5/2du1413(u+7)3/2du+49(u+7)1/2du]=67[(u+7)7/2]13845[(u+7)5/2]13+49[(u+7)3/2]13=...

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