hasil dari integral x^2/akaar x^3+2 adalah

J x^2 /√(x^3+2) dx..( J adalsh integral)
= J x^2 /(x^3+2)^1/2 dx
= J x^2 (x^3+2)^-1/2 dx
misal : u = x^3 + 2
du/dx = 3x^2
(3x^2)dx = du
dx = du/3x^2, maka
J x^2 .u^-1/2 du/3x^2
= 1/3 J u^-1/2 du
= 1/3 { (u^)-1/2+1 } / (-1/2+1) + c
= 1/3 (u^1/2) /(1/2) + c
= 2/3.u^1/2 + c
= (2/3)√u + c
= (2/3)√(x^3+2) + c

[tex]\displaystyle \text{misal:}\\x^3+2=u\\3x^2\,dx=du\\\\\int\frac{x^2}{\sqrt{x^3+2}}\,dx=\int\frac{x^2\,dx}{\sqrt{x^3+2}}\\\int\frac{x^2}{\sqrt{x^3+2}}\,dx=\int\frac{\frac13\,du}{\sqrt{u}}\\\int\frac{x^2}{\sqrt{x^3+2}}\,dx=\int\frac13u^{-\frac12}\,du\\\int\frac{x^2}{\sqrt{x^3+2}}\,dx=\frac13\cdot\frac{1}{-\frac12+1}u^{-\frac12+1}+C\\\int\frac{x^2}{\sqrt{x^3+2}}\,dx=\frac13\cdot2u^{\frac12}+C\\\boxed{\boxed{\int\frac{x^2}{\sqrt{x^3+2}}\,dx=\frac23\sqrt{x^3+2}+C}}[/tex]