Hasil integral (x+3) / ∛x²+6x dx adalah....
tolong bantu ya :D
tolong bantu ya :D
Takamori37
![$\begin{align}\int\frac{x+3}{\sqrt[3]{x^2+6x}}\, dx \\ \text{Dengan:} \\ u=x^2+6x \\ du=2x+6\, dx&\to \frac{1}{2}du=x+3\, dx \\ \text{Sehingga,} \\ \int\frac{x+3}{\sqrt[3]{x^2+6x}}\, dx&=\int\frac{\frac{1}{2}du}{\sqrt[3]{u}} \\ &=\int\frac{1}{2}u^{-1/3}\, du \\ &=\frac{1}{2}\times\frac{u^{2/3}}{^2/_3}+C \\ &=\frac{3}{4}u^{2/3}+C \\ \text{Substitusikan:} \\ \int\frac{x+3}{\sqrt[3]{x^2+6x}}\, dx&=\frac{3}{4}\sqrt[3]{(x^2+6x)^2}+C \end{align}](https://tex.z-dn.net/?f=%24%5Cbegin%7Balign%7D%5Cint%5Cfrac%7Bx%2B3%7D%7B%5Csqrt%5B3%5D%7Bx%5E2%2B6x%7D%7D%5C%2C+dx+%5C%5C+%5Ctext%7BDengan%3A%7D+%5C%5C+u%3Dx%5E2%2B6x+%5C%5C+du%3D2x%2B6%5C%2C+dx%26%5Cto+%5Cfrac%7B1%7D%7B2%7Ddu%3Dx%2B3%5C%2C+dx+%5C%5C+%5Ctext%7BSehingga%2C%7D+%5C%5C+%5Cint%5Cfrac%7Bx%2B3%7D%7B%5Csqrt%5B3%5D%7Bx%5E2%2B6x%7D%7D%5C%2C+dx%26%3D%5Cint%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7Ddu%7D%7B%5Csqrt%5B3%5D%7Bu%7D%7D+%5C%5C+%26%3D%5Cint%5Cfrac%7B1%7D%7B2%7Du%5E%7B-1%2F3%7D%5C%2C+du+%5C%5C+%26%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%5Cfrac%7Bu%5E%7B2%2F3%7D%7D%7B%5E2%2F_3%7D%2BC+%5C%5C+%26%3D%5Cfrac%7B3%7D%7B4%7Du%5E%7B2%2F3%7D%2BC+%5C%5C+%5Ctext%7BSubstitusikan%3A%7D+%5C%5C+%5Cint%5Cfrac%7Bx%2B3%7D%7B%5Csqrt%5B3%5D%7Bx%5E2%2B6x%7D%7D%5C%2C+dx%26%3D%5Cfrac%7B3%7D%7B4%7D%5Csqrt%5B3%5D%7B%28x%5E2%2B6x%29%5E2%7D%2BC+%5Cend%7Balign%7D "$\begin{align}\int\frac{x+3}{\sqrt[3]{x^2+6x}}\, dx \\ \text{Dengan:} \\ u=x^2+6x \\ du=2x+6\, dx&\to \frac{1}{2}du=x+3\, dx \\ \text{Sehingga,} \\ \int\frac{x+3}{\sqrt[3]{x^2+6x}}\, dx&=\int\frac{\frac{1}{2}du}{\sqrt[3]{u}} \\ &=\int\frac{1}{2}u^{-1/3}\, du \\ &=\frac{1}{2}\times\frac{u^{2/3}}{^2/_3}+C \\ &=\frac{3}{4}u^{2/3}+C \\ \text{Substitusikan:} \\ \int\frac{x+3}{\sqrt[3]{x^2+6x}}\, dx&=\frac{3}{4}\sqrt[3]{(x^2+6x)^2}+C \end{align}")
2 votes
Thanks 1
Irene22
wah... makasih banyak kak :D ^_^
Takamori37
Iya, sama-sama.
