Question

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[tex]∫ \frac{dx}{ \sqrt{16 + 25x - 4 {x}^{2} } } [/tex]
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Answer 1

[tex]\large\text{$\begin{aligned}\frac{1}{2}\arcsin\left(\frac{8x-25}{\sqrt{881}}\right)+C\end{aligned}$}[/tex] PembahasanIntegral[tex]\begin{aligned}&\int\frac{dx}{\sqrt{16+25x-4{x}^2}}\\&{=\ }\int\frac{1}{\sqrt{16+25x-4{x}^2}}\,dx\end{aligned}[/tex]Pada penyebut, terdapat bentuk akar kuadrat yang memuat fungsi kuadrat. Kita lengkapkan menjadi bentuk yang melibatkan kuadrat sempurna.[tex]\begin{aligned}&16+25x-4x^2\\&{=\ }-4x^2+25x+16\\&{=\ }-4\left(x^2-\frac{25}{4}x-4\right)\\&{=\ }-4\left[x^2-\frac{25}{4}x+\left(-\frac{25}{8}\right)^2-4-\left(-\frac{25}{8}\right)^2\right]\\&{=\ }-4\left[\left(x^2-\frac{25}{8}\right)^2-4-\frac{625}{64}\right]\\&{=\ }-4\left[\left(x-\frac{25}{8}\right)^2-\frac{256+625}{64}\right]\\&{=\ }-4\left[\left(x-\frac{25}{8}\right)^2-\frac{881}{64}\right]\end{aligned}[/tex][tex]\begin{aligned}&{=\ }\frac{881}{16}-4\left(x-\frac{25}{8}\right)^2\\&{=\ }\frac{881}{16}-2^2\left(x-\frac{25}{8}\right)^2\\&{=\ }\frac{881}{16}-\left[2\left(x-\frac{25}{8}\right)\right]^2\\&{=\ }\frac{881}{16}-\left(2x-\frac{25}{4}\right)^2\\&{=\ }\frac{881}{16}-\left(\frac{8x-25}{4}\right)^2\\\end{aligned}[/tex]Maka:[tex]\begin{aligned}&\int\frac{dx}{\sqrt{16+25x-4{x}^2}}\\\\&{=\ }\int\frac{1}{\sqrt{\dfrac{881}{16}-\left(\dfrac{8x-25}{4}\right)^2}}\,dx\quad...(\star)\end{aligned}[/tex]Selanjutnya, kita akan menggunakan integral substitusi.[tex]\begin{aligned}&{\sf Ambil\ }u\ {\sf sedemikian\ hingga}\\&\frac{8x-25}{4}=u\sqrt{\frac{881}{16}}=\frac{u\sqrt{881}}{4}\\&{\Rightarrow\ }u=\frac{8x-25}{\sqrt{881}}\\&{\Rightarrow\ }\frac{du}{dx}=\frac{8}{\sqrt{881}}\\&{\therefore\ \ }dx=\frac{\sqrt{881}}{8}\,du\\&{\therefore\ \ }\left(\frac{8x-25}{4}\right)^2=\frac{881u^2}{16}\end{aligned}[/tex]Substitusi ke dalam [tex](\star)[/tex] :[tex]\begin{aligned}&{(\star)\to\ }\int\frac{1}{\sqrt{\dfrac{881}{16}-\left(\dfrac{8x-25}{4}\right)^2}}\,dx\\&{=\ }\int\frac{1}{\sqrt{\dfrac{881}{16}-\dfrac{881u^2}{16}}}\cdot\frac{\sqrt{881}}{8}\,du\\&{=\ }\int\frac{\sqrt{881}}{8\sqrt{\dfrac{881}{16}-\dfrac{881u^2}{16}}}\,du\\&{=\ }\int\frac{\sqrt{881}}{\dfrac{8}{4}\sqrt{881-881{u}^2}}\,du\\&{=\ }\int\frac{\cancel{\sqrt{881}}}{2\cancel{\sqrt{881}}\sqrt{1-u^2}}\,du\\&{=\ }\int\frac{1}{2\sqrt{1-u^2}}\,du\end{aligned}[/tex][tex]\begin{aligned}&{=\ }\frac{1}{2}\int\frac{1}{\sqrt{1-u^2}}\,du\quad...(\star\star)\\\end{aligned}[/tex]Bentuk terakhir yang diperoleh memuat:[tex]\displaystyle\int\frac{1}{\sqrt{1-u^2}}\,du=\arcsin(u)[/tex]Oleh karena itu:[tex]\begin{aligned}&{(\star\star)\to\ }\frac{1}{2}\int\frac{1}{\sqrt{1-u^2}}\,du\\&{=\ }\frac{1}{2}\arcsin(u)+C\\&{=\ }\frac{1}{2}\arcsin\left(\frac{8x-25}{\sqrt{881}}\right)+C\qquad\blacksquare\\\end{aligned}[/tex] KESIMPULAN[tex]\begin{aligned}&\int\frac{dx}{\sqrt{16+25x-4{x}^2}}\\\\&{=\ }\boxed{\ \frac{1}{2}\arcsin\left(\frac{8x-25}{\sqrt{881}}\right)+C\ }\end{aligned}[/tex]