∫x⁵(2-x³)pangkat 1/2 dx= Mohon bantuannya dijawab ya segera... Semoga kebaikan saudara dibalas Allah SWT dengan kebaikan pula....

[tex]\displaystyle \text{misal :}\\x^3=u\\3x^2\,dx=du\\\\\int x^5\sqrt{2-x^3}\,dx=\int \sqrt{2-x^3}\cdot x^3\cdot\frac13(3x^2)\,dx\\\int x^5\sqrt{2-x^3}\,dx=\int \sqrt{2-u}\cdot u\cdot\frac13\,du\\\int x^5\sqrt{2-x^3}\,dx=\frac13\int\sqrt{2-u}\cdot u\,du\\\int x^5\sqrt{2-x^3}\,dx=\frac13\int\sqrt{2-u}(-(2-u)+2)\,du\\\int x^5\sqrt{2-x^3}\,dx=\frac13\int2(2-u)^\frac12-(2-u)^\frac32\,du\\\int x^5\sqrt{2-x^3}\,dx=\frac13\left(2\cdot\frac23(2-u)^\frac32-\frac25(2-u)^\frac52\right)+C[/tex]
[tex]\displaystyle \int x^5\sqrt{2-x^3}\,dx=\frac49(2-u)^\frac32-\frac2{15}(2-u)^\frac52+C\\\boxed{\boxed{\int x^5\sqrt{2-x^3}\,dx=\frac49(2-x^3)^\frac32-\frac2{15}(2-x^3)^\frac52+C}}[/tex]

∫ x^5 (2 - x^3)^1/2 dx
= ∫ x^2 . x^3 (1 - x^3)^1/2 dx
Misal
u = x^3
du = 3x^2 dx
dx = du/(3x^2)
= ∫ x^2 . u . (1 - u)^1/2 du/(3x^2)
= ∫ 1/3 u (1 - u)^1/2 du
Gunakan integral parsial dengan cara tabel
Turunan ... Integral
(+) 1/3 u ....... (1 - u)^1/2
(-) 1/3 .......... -2/3 (1 - u)^3/2
(+) 0 ............ -2/3 . -2/5 (1 - u)^5/2

= 1/3 u (-2/3 (1 - u)^3/2 - 1/3 (-2/3 . -1/5 (1 - u)^5/2) + C
= -2/9 u (1 - u)^3/2 - 2/45 (1 - u)^5/2 + C
= -2/9 x^3 (1 - x^3)^3/2 - 2/45 (1 - x^3)^5/2 + C

= -2/9 x^3 (1 - x^3) √(1 - x^3) - 2/45 (1 - x^3)^2 √(1 - x^3) + C

^ = pangkat