[tex] \displaystyle \sf \int \frac{45yz \times 5xy \times \sqrt{x + 12y} }{ \sqrt{x} \times 2x + 12x.... Question from @misrokhah439

July 2022 1 20 Report

[tex] \displaystyle \sf \int \frac{45yz \times 5xy \times \sqrt{x + 12y} }{ \sqrt{x} \times 2x + 12xz} \: dz[/tex]

nt: bantu Kaks otak sy lagi leg :\

henriyulianto

$\begin{aligned}&\int\frac{45yz\times 5xy\times\sqrt{x+12y}}{\sqrt{x}\times2x+12xz}\,dz\\&=\ \boxed{\ \bf\frac{25y^2\sqrt{x+12y}\left[6z-\sqrt{x}\ln\left(\left|\sqrt{x}+6z\right|\right)\right]}{8}+C\ }\end{aligned}$

Pembahasan

Integral

$\begin{aligned}&\int\frac{45yz\times 5xy\times\sqrt{x+12y}}{\sqrt{x}\times2x+12xz}\,dz\\&{=\ }\int\frac{\left(225\cancel{x}y^2\sqrt{x+12y}\right)\times z}{2\cancel{x}\left(\sqrt{x}+6z\right)}\,dz\\&{=\ }\int\frac{\left(225y^2\sqrt{x+12y}\right)\times z}{2\left(\sqrt{x}+6z\right)}\,dz\\\\&{\quad}...\ \textsf{keluarkan konstanta}\end{aligned}$
 

$\begin{aligned}&{=\ }\frac{225y^2\sqrt{x+12y}}{2}\cdot\int\frac{z}{\sqrt{x}+6z}\,dz\\\\&{\quad}\left[\ \begin{aligned}&{\sf Ambil\ }u=\sqrt{x}+6z\\&\Rightarrow du=6dz\\&\Rightarrow dz=\frac{1}{6}du\\&\Rightarrow z=\frac{u-\sqrt{x}}{6}\end{aligned}\right.\\\\&{=\ }\frac{225y^2\sqrt{x+12y}}{2}\cdot\int\frac{\left(\frac{u-\sqrt{x}}{6}\right)}{u}\cdot\frac{1}{6}du\\&{=\ }\frac{225y^2\sqrt{x+12y}}{2}\cdot\int\frac{u-\sqrt{x}}{36u}\,du\end{aligned}$

$\begin{aligned}&{=\ }\frac{225y^2\sqrt{x+12y}}{72}\cdot\int\frac{u-\sqrt{x}}{u}\,du\\&{=\ }\frac{\cancel{9}\cdot25y^2\sqrt{x+12y}}{\cancel{9}\cdot8}\cdot\int\left(1-\frac{\sqrt{x}}{u}\right)du\\&{=\ }\frac{25y^2\sqrt{x+12y}}{8}\cdot\left(\int1\,du-\sqrt{x}\int\frac{1}{u}\,du\right)\\&{=\ }\frac{25y^2\sqrt{x+12y}}{8}\cdot\left[u-\sqrt{x}\ln\left(\left|u\right|\right)\right]\\\\&{\quad}...\ \textsf{substitusi kembali } u\end{aligned}$
 

$\begin{aligned}&{=\ }\frac{25y^2\sqrt{x+12y}}{8}\cdot\left[\sqrt{x}+6z-\sqrt{x}\ln\left(\left|\sqrt{x}+6z\right|\right)\right]+C\\&{=\ }\frac{25y^2\sqrt{x+12y}\cdot\sqrt{x}}{8}\\&{\quad}+\frac{25y^2\sqrt{x+12y}}{8}\cdot\left[6z-\sqrt{x}\ln\left(\left|\sqrt{x}+6z\right|\right)\right]\\&{\quad}+C\end{aligned}$
 

$\begin{aligned}&{\quad}\left[\ \begin{aligned}&\textsf{Perhatikan bahwa }\frac{25y^2\sqrt{x+12y}\cdot\sqrt{x}}{8}\\&\textsf{adalah konstanta dan tidak memuat }z,\\&\textsf{sehingga dapat "dilebur" dengan }C.\\\end{aligned}\right.\\\\&{=\ }\frac{25y^2\sqrt{x+12y}}{8}\cdot\left[6z-\sqrt{x}\ln\left(\left|\sqrt{x}+6z\right|\right)\right]+C\\&{=\ }\boxed{\ \bf\frac{25y^2\sqrt{x+12y}\left[6z-\sqrt{x}\ln\left(\left|\sqrt{x}+6z\right|\right)\right]}{8}+C\ }\end{aligned}$