1. hasil dari integral x^2 akar 2x^3 - 1 dx adalah 2. jika f(n) = 2^n+2 6^n-4 dan g(n) = 12^n-1 , n bilangan asli, maka f(n) / g(n) adalah

1) misal :
u = 2x^3 - 1
du/dx = 6x^2 => dx = du/(6x^2)
Integral (x^2 √(2x^3 - 1)) dx
= Integral (x^2 √u ) du/(6x^2)
= Integral (1/6) u^(1/2) di
= (1/6)/(3/2) u^(3/2) + C
= 2/18 u^(3/2) + C
= (1/9) (2x^3 - 1)^3/2 + C

= (1/9) √(2x^3 - 1)^3 + C

= (1/9) (2x^3 - 1) √(2x^3 - 1) + C

2) f(n) / g(n)
= (2^(n + 2) . 6^(n - 4)) / (12^(n - 1))
= (2^(n + 2) . 6^(n - 4)) / (2^(n - 1) . 6^(n - 1)
= 2^(n + 2 - (n - 1)) . 6^(n - 4 - (n - 1))
= 2^3 . 6^(-3)
= 2^3 . 1/(6^3)
= (2/6)^3
= (1/3)^3
= 1/27

[tex]\displaystyle \text{misal :}\\2x^3-1=u\\6x^2\,dx=du\\\\\int x^2\sqrt{2x^3-1}\,dx=\int \sqrt{2x^3-1}x^2\,dx\\\int x^2\sqrt{2x^3-1}\,dx=\int \sqrt{u}\cdot\frac16\,du\\\int x^2\sqrt{2x^3-1}\,dx=\int\frac16u^\frac12\,du\\\int x^2\sqrt{2x^3-1}\,dx=\frac16\cdot\frac23u^\frac32+C\\\int x^2\sqrt{2x^3-1}\,dx=\frac19u^\frac32+C\\\boxed{\boxed{\int x^2\sqrt{2x^3-1}\,dx=\frac19(2x^3-1)^\frac32+C}}[/tex]

[tex]\displaystyle \frac{f(n)}{g(n)}=\frac{2^{n+2}\cdot6^{n-4}}{12^{n-1}}\\\frac{f(n)}{g(n)}=\frac{2^{n+2}\cdot2^{n-4}\cdot3^{n-4}}{4^{n-1}\cdot3^{n-1}}\\\frac{f(n)}{g(n)}=\frac{2^{2n-2}\cdot3^{n-4-n+1}}{2^{2n-2}}\\\frac{f(n)}{g(n)}=3^{-3}\\\boxed{\boxed{\frac{f(n)}{g(n)}=\frac1{27}}}[/tex]