Bantuin jawab dgn caranya,,
Hasil
2x
dx = ....
A.
( 8 - 
) + C
B.
( 8 - ) + C
C.
( 8 - ) + C
D.
( 8 - ) + C
E.
( 8 - ) + C
Hasil
A.
B.
C.
D.
E.
Takamori37
![\int 2x\sqrt{8-x^2}\, dx \\ $Anggap:$ \\ u=8-x^2 \\ du=-2x\, dx ~~\to~~2x=-\, du \\ $Sehingga:$ \\ \int -\sqrt{u}\, du \\ -\int u^{1/2}\, du=\displaystyle -\frac{u^{3/2}}{3/2}+C=-\frac{2}{3}u^{3/2}+C \\\\ -\frac{2}{3}u\sqrt{u}+C \\ \text{Substitusi ulang $u$, didapat:} \\ \boxed{\int2x\sqrt{8-x^2}\, dx=-\frac{2}{3}(8-x^2)\sqrt{8-x^2}+C\;\;\;[E]}](https://tex.z-dn.net/?f=%5Cint+2x%5Csqrt%7B8-x%5E2%7D%5C%2C+dx+%5C%5C+%24Anggap%3A%24+%5C%5C+u%3D8-x%5E2+%5C%5C+du%3D-2x%5C%2C+dx+%7E%7E%5Cto%7E%7E2x%3D-%5C%2C+du+%5C%5C+%24Sehingga%3A%24+%5C%5C+%5Cint+-%5Csqrt%7Bu%7D%5C%2C+du+%5C%5C+-%5Cint+u%5E%7B1%2F2%7D%5C%2C+du%3D%5Cdisplaystyle+-%5Cfrac%7Bu%5E%7B3%2F2%7D%7D%7B3%2F2%7D%2BC%3D-%5Cfrac%7B2%7D%7B3%7Du%5E%7B3%2F2%7D%2BC+%5C%5C%5C%5C+-%5Cfrac%7B2%7D%7B3%7Du%5Csqrt%7Bu%7D%2BC+%5C%5C+%5Ctext%7BSubstitusi+ulang+%24u%24%2C+didapat%3A%7D+%5C%5C+%5Cboxed%7B%5Cint2x%5Csqrt%7B8-x%5E2%7D%5C%2C+dx%3D-%5Cfrac%7B2%7D%7B3%7D%288-x%5E2%29%5Csqrt%7B8-x%5E2%7D%2BC%5C%3B%5C%3B%5C%3B%5BE%5D%7D "\int 2x\sqrt{8-x^2}\, dx \\ $Anggap:$ \\ u=8-x^2 \\ du=-2x\, dx ~~\to~~2x=-\, du \\ $Sehingga:$ \\ \int -\sqrt{u}\, du \\ -\int u^{1/2}\, du=\displaystyle -\frac{u^{3/2}}{3/2}+C=-\frac{2}{3}u^{3/2}+C \\\\ -\frac{2}{3}u\sqrt{u}+C \\ \text{Substitusi ulang $u$, didapat:} \\ \boxed{\int2x\sqrt{8-x^2}\, dx=-\frac{2}{3}(8-x^2)\sqrt{8-x^2}+C\;\;\;[E]}")
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