Quiz last (Off sementara) (1/15)

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$intlimitsfrac{1}{2xsqrt{x}},dx=?$
a. $frac{1}{sqrt{-x} } +C$
b. $frac{-1}{sqrt{-x} } +C$
c. $frac{1}{sqrt{x} } +C$
d. $frac{-1}{sqrt{x}} +C$
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Hint :
Gunakan rumus $intlimits(kx^{n})dx=k(frac{1}{n+1} )x^{n+1}+C,nneq -1$
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Note : Alasan ada di akhir kuis…..

Quiz last (Off sementara) (1/15)

Jawaban:

integral tak tentu

…

ʃ 1/(2x√x) dx

ʃ 1/ (2 . x√x) dx

ʃ 1/2 . 1/x√x dx

½ ʃ ¹/x^3/2 dx

½ ʃ x^-3/2 dx

½ ʃ x^(-3/2 + 1) /(-3/2+1) dx

½ ʃ x^(-1/2)/ (-1/2) dx

ʃ (1/2) . 1/(-1/2) . x^-1/2 dx

= (1/2) . (-2) . x^-1/2 + C

= -1 . x^-1/2 + c

= -1/x^½ + C

= -1/√x + C

Penjelasan dengan langkah-langkah:

$intlimitsfrac{1}{2xsqrt{x}},dx \ ∫ frac{1}{2 {x}^{ frac{3}{2} } } dx \ ∫ frac{1}{2} {x}^{ - frac{3}{2} } dx \ frac{1}{2} times frac{ {x}^{ - frac{3}{2} + 1 } }{} \ frac{1}{2} times {x}^{ - frac{1}{2} } \ frac{1}{2} times frac{ {x}^{ - frac{1}{2} } }{ - frac{1}{2} } \ - frac{1}{ {x}^{ frac{1}{2} } } \ = - frac{1}{ sqrt{x} } + C$