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misal: u = √x
du = 1/(2√x) dx
∫ 2^(√x) dx = 2.∫ (2^(u)).u du
= (2^(u+1))(u.ln (2) - 1)/ln² (2)
= (2^(1+√x)).[(√x).ln (2) - 1]/ln² (2)
2. ∫ x².5^(x²) dx = susahh euyy, lieuurrr
3. ∫ x².cos 3x dx
misal:
u = x²
du = 2x dx
dv = cos 3x dx
v = (1/3).sin 3x
∫ x².cos 3x dx = u.v - ∫ vdu
= (x²/3).sin 3x - ∫ (2x/3).sin 3x dx
= (x²/3).sin 3x + (2x/9).cos 3x - ∫ (2/9).cos 3x dx
= (x²/3).sin 3x + (2x/9).cos 3x - (2/27).sin 3x + C
= (1/27).((9x²-2).sin 3x + 6x.cos 3x) + C
4. 3ˣ.ln (x+5) ← ane ganti z pake "x"
y = u.v
y' = u.v' + u'.v
Dx[3ˣ.ln (x+5)] = 3ˣ.(1/(x+5)) + (3ˣ.ln (3)).(ln (x+5))
= 3ˣ.(1/(x+5) + ln (3).ln (x+5))
5. Dx[6^(x²)] = 6.(2x).(6^(x²-1)).ln (6)
= 2x.(6^(x²)).ln (6)
6. ∫ x2^(x²) dx
misal:
u = x²
du = 2x dx
subtitusi x² dan dx:
(1/2).∫ 2^u du = (2^(u+1))/4.ln (2) + C
= [(2^(x²-1))/ln 2] + C