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ikut sampai sini

-2∫ u²/(u²+1)³ du,

saya lanjutkan u = tan(t) , du = sec²(t) dt

-2∫(1/2 sin(2t))² dt = (sin(4t)-4t)/16

=sin(4arctan(u))/16 - arctan(u) / 4

=sin(4arctan(√ ((1/x) - 1) ))/16 - arctan(√ ((1/x) - 1) ) / 4

=limit t→0⁺[sin(4arctan(√ ((1/x) - 1) ))/16 - arctan(√ ((1/x) - 1) ) / 4] dari t sampai 1

=[(0-0)-(0-pi/8)]

=pi/8

Jangan pakai substitusi u yang kamu pakai itu, langkah pertama sederhanakan dahulu soal itu:

∫ [0,1] x √ ((1/x) - 1) dx

= ∫ (0,1) x √(1-x)/x dx

kalikan dengan √x/√x, sehingga:

= ∫ (0,1) x*√x(1-x)/ x dx

= ∫ (0,1) √x(1-x) dx

Gunakan cara melengkapkan untuk persamaan di dalam kurung:

x(1-x)= x-x^2= -(x^2-x)= -(x^2-x+1/4 -1/4)= -((x-1/2)^2 -1/4)= (1/4 -(x-1/2)^2)

Persamaan terakhir berubah menjadi:

= ∫ (0,1) √(1/4)-(x-1/2)^2 dx

Dari sini, kkita gunakan integral tak tentu dulu:

Misal p=x-1/2

dp=dx

= ∫ √ 1/4 -p^2 dp

= ∫ √(1-4p^2)/4 dp

= 1/2∫ √(1-(2p)^2 dp

Misal 2p= cos t

p= cos t/2

dp= -sin t/2 dt, maka:

= 1/2∫ √(1-cos^2 t). -sin t/2 dt

= 1/4 ∫ sin t.(-sin t) dt

= 1/4 ∫ -sin^2 t dt

= -1/4 ∫ sin^2 t dt

Ingat cos 2t= cos^2 t-sin^2 t= 1-2sin^2 t

sin^2 t= (1-cos 2t)/2

= -1/4∫ (1-cos 2t)/2 dt

= -1/8∫ 1-cos 2t dt

= -1/8∫ 1 dt +1/8∫ cos 2t dt

= -t/8 +1/8∫ cos 2t dt

Misal y= 2t, maka dy= 2dt

= -t/8 +1/16∫ 2cos 2t dt

= -t/8 +1/16∫ cos y dy

= -t/8+1/16*(sin y) +C

Sekarang kembalikan nilai-nilainya:

=-t/8 +1/16.sin(2t) +C

2p= cos t--> t= arccos(2p), maka:

= -arccos(2p)/8 +1/16.sin(2.(arccos(2p)) +C

= -arccos(2(x-1/2))/8 +1/16.sin(2(arccos(2(x-1/2)))) +C

Sekarang, kita hitung integral tentunya:

= -arccos(2(x-1/2))/8 +1/16.sin(2(arccos(2(x-1/2)))) (dari 0 sampai 1)

= -arccos(2.1/2)/8 +1/16sin(2(arccos(2.1/2))-(-arccos(2.-1/2)/8 +1/16.sin(2arccos(2.-1/2)))

= -arccos(1)/8 +sin(2.arccos 1)/16 -(-arccos(-1)/8 +sin(2arccos(-1))/16)

= 0+0-(-pi/8 +sin(2pi)/16)

=-(-pi/8 +0)

= pi/8