Mohon bantuan caranya (Substitusi)
1.
1.
Takamori37
Fragmentasi:
![$\begin{align}\int\frac{\sqrt{x}}{1+\sqrt{x}}\, dx&=\int 1-\frac{1}{1+\sqrt{x}}\, dx \\ &=x-\int\frac{1}{\sqrt{x}+1}\, dx \\ &----------\\ &\text{Untuk }\int\frac{1}{\sqrt{x}+1}\, dx \\ &u=\sqrt{x} \\ &x=u^2 \\ &\frac{d}{du}x=2u \\ &dx=2u\, du \\ &\text{Sehingga, menjadi:} \\ &=2\int\frac{u\, du}{u+1} \\ &=2\int\left[1-\frac{1}{u+1}\right]\, du \\ &=2(u-\ln|u+1|)+C \\ &\text{Substitusikan:} \\ &=2\sqrt{x}-2\ln|\sqrt{x}+1|+C \\ &- - - - - - - - - - \\ &\text{Sehingga,}\end{align}](https://tex.z-dn.net/?f=%24%5Cbegin%7Balign%7D%5Cint%5Cfrac%7B%5Csqrt%7Bx%7D%7D%7B1%2B%5Csqrt%7Bx%7D%7D%5C%2C+dx%26%3D%5Cint+1-%5Cfrac%7B1%7D%7B1%2B%5Csqrt%7Bx%7D%7D%5C%2C+dx+%5C%5C+%26%3Dx-%5Cint%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%7D%2B1%7D%5C%2C+dx+%5C%5C+%26----------%5C%5C+%26%5Ctext%7BUntuk+%7D%5Cint%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%7D%2B1%7D%5C%2C+dx+%5C%5C+%26u%3D%5Csqrt%7Bx%7D+%5C%5C+%26x%3Du%5E2+%5C%5C+%26%5Cfrac%7Bd%7D%7Bdu%7Dx%3D2u+%5C%5C+%26dx%3D2u%5C%2C+du+%5C%5C+%26%5Ctext%7BSehingga%2C+menjadi%3A%7D+%5C%5C+%26%3D2%5Cint%5Cfrac%7Bu%5C%2C+du%7D%7Bu%2B1%7D+%5C%5C+%26%3D2%5Cint%5Cleft%5B1-%5Cfrac%7B1%7D%7Bu%2B1%7D%5Cright%5D%5C%2C+du+%5C%5C+%26%3D2%28u-%5Cln%7Cu%2B1%7C%29%2BC+%5C%5C+%26%5Ctext%7BSubstitusikan%3A%7D+%5C%5C+%26%3D2%5Csqrt%7Bx%7D-2%5Cln%7C%5Csqrt%7Bx%7D%2B1%7C%2BC+%5C%5C+%26-+-+-+-+-+-+-+-+-+-+%5C%5C+%26%5Ctext%7BSehingga%2C%7D%5Cend%7Balign%7D "$\begin{align}\int\frac{\sqrt{x}}{1+\sqrt{x}}\, dx&=\int 1-\frac{1}{1+\sqrt{x}}\, dx \\ &=x-\int\frac{1}{\sqrt{x}+1}\, dx \\ &----------\\ &\text{Untuk }\int\frac{1}{\sqrt{x}+1}\, dx \\ &u=\sqrt{x} \\ &x=u^2 \\ &\frac{d}{du}x=2u \\ &dx=2u\, du \\ &\text{Sehingga, menjadi:} \\ &=2\int\frac{u\, du}{u+1} \\ &=2\int\left[1-\frac{1}{u+1}\right]\, du \\ &=2(u-\ln|u+1|)+C \\ &\text{Substitusikan:} \\ &=2\sqrt{x}-2\ln|\sqrt{x}+1|+C \\ &- - - - - - - - - - \\ &\text{Sehingga,}\end{align}")
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Risep
makasih :D
Risep
satunya dong please
Takamori37
Yang arctan aku suka pusing kadang-kadang.
Coba dulu, ya.
Risep
yes
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