Quiz || 2 - 5 { Easy }Tentukan hasil integral

Berikut ini adalah pertanyaan dari unknown pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Quiz || 2 - 5 { Easy }Tentukan hasil integral tak tentu fungsi berikut :
$\\$
$\int {\frac{x^2+1}{x^4+1}}$

:- Gampang :-;
:- Okee Good luck :-:

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Hasil dari $\displaystyle{\int\limits {\frac{x^2+1}{x^4+1}} \, dx }$ adalah $\displaystyle{\boldsymbol{\frac{arctan\left ( \sqrt{2}x+1 \right )+arctan\left ( \sqrt{2}x-1 \right )}{\sqrt{2}}+C} }$ .

PEMBAHASAN

Integral merupakan operasi yang menjadi kebalikan dari operasi turunan/diferensial. Sehingga integral sering juga disebut sebagai antiturunan

$\displaystyle{f(x)=\int\limits {\left [ \frac{df(x)}{dx} \right ]} \, dx}$

Untuk fungsi invers trigonometri :

$\displaystyle{\int\limits {\frac{1}{\sqrt{1-x^2}}} \, dx=arcsinx+C }$

$\displaystyle{\int\limits {\frac{1}{x^2+1}} \, dx=arctanx+C }$

DIKETAHUI

$\displaystyle{\int\limits {\frac{x^2+1}{x^4+1}} \, dx= }$

DITANYA

Tentukan hasilnya.

PENYELESAIAN

$\displaystyle{\int\limits {\frac{x^2+1}{x^4+1}} \, dx }$

$\displaystyle{=\int\limits {\frac{x^2+1}{x^4+2x^2+1-2x^2}} \, dx }$

$\displaystyle{=\int\limits {\frac{x^2+1}{(x^2+1)^2-(\sqrt{2}x)^2}} \, dx }$

$\displaystyle{=\int\limits {\frac{x^2+1}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}} \, dx }$

$---------------$

Misal :

$\displaystyle{\frac{x^2+1}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}=\frac{A}{(x^2+\sqrt{2}x+1)}+\frac{B}{(x^2-\sqrt{2}x+1)} }$

$\displaystyle{\frac{x^2+1}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}=\frac{A(x^2-\sqrt{2}x+1)+B(x^2+\sqrt{2}x+1)}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)} }$

$\displaystyle{\frac{x^2+1}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}=\frac{(A+B)x^2+(-\sqrt{2}A+\sqrt{2}B)x+A+B}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)} }$

Cari nilai A dan B dengan menyamakan kedua ruas.

Koefisien x² :

$A+B=1$

$B=1-A~~~...(i)$

Koefisien x :

$-\sqrt{2}A+\sqrt{2}B=0~~~...kedua~ruas~dibagi~\sqrt{2}$

$-A+B=0~~~...substitusi~pers.(i)$

$-A+1-A=0$

$2A=1$

$\displaystyle{A=\frac{1}{2}~\to~B=1-\frac{1}{2}=\frac{1}{2} }$

Diperoleh :

$\displaystyle{\frac{x^2+1}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}=\frac{1}{2(x^2+\sqrt{2}x+1)}+\frac{1}{2(x^2-\sqrt{2}x+1)} }$

$---------------$

$\displaystyle{=\int\limits {\left [ \frac{1}{2(x^2+\sqrt{2}x+1)}+\frac{1}{2(x^2-\sqrt{2}x+1)} \right ]} \, dx }$

$\displaystyle{=\frac{1}{2}\int\limits {\left [ \frac{1}{x^2+\sqrt{2}x+1}+\frac{1}{x^2-\sqrt{2}x+1} \right ]} \, dx }$

$\displaystyle{=\frac{1}{2}\int\limits {\left [ \frac{1}{(x^2+\sqrt{2}x+\frac{1}{2})+(1-\frac{1}{2})}+\frac{1}{(x^2-\sqrt{2}x+\frac{1}{2})+(1-\frac{1}{2})} \right ]} \, dx }$

$\displaystyle{=\frac{1}{2}\int\limits {\left [ \frac{1}{(x+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}+\frac{1}{(x-\frac{\sqrt{2}}{2})^2+\frac{1}{2}} \right ]} \, dx }$

$---------------$

Untuk $\displaystyle{\int\limits {\frac{1}{(x+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}} \, dx }$ :

Misal :

$\displaystyle{u=x+\frac{\sqrt{2}}{2}~\to~du=dx }$

Maka :

$\displaystyle{\int\limits {\frac{1}{(x+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}} \, dx }$

$\displaystyle{=\int\limits {\frac{1}{u^2+\frac{1}{2}}} \, du }$

$\displaystyle{=\int\limits {\frac{1}{\frac{2u^2+1}{2}}} \, du }$

$\displaystyle{=\int\limits {\frac{2}{2u^2+1}} \, du }$

$\displaystyle{=\int\limits {\frac{2}{(\sqrt{2}u)^2+1}} \, du }$

$.~~~~~~~~~~~~~~Misal:~v=\sqrt{2}u~\to~dv=\sqrt{2}du$

$\displaystyle{=\int\limits {\frac{2}{v^2+1}} \, \frac{dv}{\sqrt{2}} }$

$\displaystyle{=\frac{2}{\sqrt{2}}\int\limits {\frac{1}{v^2+1}} \, dv }$

$\displaystyle{=\frac{2}{\sqrt{2}}arctanv+C_1 }$

$\displaystyle{=\frac{2}{\sqrt{2}}arctan(\sqrt{2}u)+C_1 }$

$\displaystyle{=\frac{2}{\sqrt{2}}arctan\left ( \sqrt{2}x+1 \right )+C_1 }$

Untuk $\displaystyle{\int\limits {\frac{1}{(x-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}} \, dx }$ :

Misal :

$\displaystyle{u=x-\frac{\sqrt{2}}{2}~\to~du=dx }$

Maka :

$\displaystyle{\int\limits {\frac{1}{(x-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}} \, dx }$

$\displaystyle{=\int\limits {\frac{1}{u^2+\frac{1}{2}}} \, du }$

Hasilnya akan sama dengan integral sebelumnya, yaitu :

$\displaystyle{=\frac{2}{\sqrt{2}}arctan\left ( \sqrt{2}x-1 \right )+C_2 }$

$---------------$

$\displaystyle{=\frac{1}{2}\int\limits {\left [ \frac{1}{(x+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}+\frac{1}{(x-\frac{\sqrt{2}}{2})^2+\frac{1}{2}} \right ]} \, dx }$

$\displaystyle{=\frac{1}{2}\left [ \frac{2}{\sqrt{2}}arctan\left ( \sqrt{2}x+1 \right )+C_1+\frac{2}{\sqrt{2}}arctan\left ( \sqrt{2}x-1 \right )+C_2 \right ] }$