Question

Bantu jawab pakai cara dan rumus​ kak

Answer 1

Penjelasan dengan langkah-langkah:

9. ∫ 3/(2x - 2) dx

u = 2x - 2

du/dx = 2

dx = 1/2 du

∫ 3/(2x - 2) dx

= 3/u . 1/2 du

= 3/2u du

= 3/2 . 1/u du

= 3/2 . ∫ 1/u du

= 3/2 . ln |2x - 2| + c

10. ∫ 6x/(√3x² + 2) dx

∫ 6x(3x² + 2)^-1/2 dx

u = 3x² + 2

du/dx = 6x

dx = 1/6x du

∫ 6x/(√3x² + 2) dx

= 6x . u^-1/2 . 1/6x du

= u^-1/2 du

= 1/(-1/2 + 1) u^1/2 + c

= 1/(1/2) . (3x² + 2)^1/2 + c

= 2 . √3x² + 2 + c

= 2√(3x² + 2) + c

∫ 1/u du = In |u| + c

∫ ax^n = 1/(n + 1) . ax^(n + 1) + c

Answer 2

9. Misalkan x - 1 = u

[tex]\sf\int {\frac{3}{(2x-2)} } \, dx \\\\=3\int {\frac{1}{(2x-2)} } \, dx\\\\=3\int {\frac{1}{2u} } \, du\\\\=3\cdot \frac{1}{2} \int {\frac{1}{u} } \, du\\\\=\frac{3}{2} \int {\frac{1}{u} } \, du\\\\=\frac{3}{2} In|u|+C\\\\\boxed{\sf=\frac{3}{2} In|x-1|+C}[/tex]

10. Misalkan 3x² + 2 = u

[tex]\sf\int {\frac{6x}{\sqrt{3x^{2} +2} } } \, dx\\\\=6\cdot \int {\frac{x}{\sqrt{3x^{2} +2} } } \, dx\\\\=6\cdot \int {\frac{1}{6\sqrt{u} } } \, dx\\\\=6\cdot \frac{1}{6} \int {\frac{1}{\sqrt{u} } } \, dx\\\\= \int {\frac{1}{\sqrt{u} } } \, dx\\\\= \int {u^{-\frac{1}{2} } } \, dx\\\\=\frac{1}{1+(-\frac{1}{2}) } u^{1+(-\frac{1}{2}) } +C\\\\=\frac{1}{\frac{1}{2} } u^{\frac{1}{2} } +C\\\\=2 u^{\frac{1}{2} } +C\\\\=2 (3x^{2} +2)^{\frac{1}{2} } +C\\\\\boxed{\sf=2\sqrt{3x^{2} +2} +C}[/tex]

[tex]\boxed{\sf{shf}}[/tex]