QUIZ #6, hasil dari ,

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QUIZ #6

hasil dari  \displaystyle \int \frac{1}{x^4 + 1} \: \text{dx}

Hasil dari \displaystyle{\int\limits {\frac{1}{x^4+1}} \, dx} adalah \boldsymbol{\frac{1}{4\sqrt{2}}ln\left | \frac{2x^2+2\sqrt{2}x+2}{2x^2-2\sqrt{2}x+2} \right |+\frac{1}{2\sqrt{2}}\left  tan^{-1}(\sqrt{2}x+1)+tan^{-1}(\sqrt{2}x-1) \right +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}

Sifat – sifat operasi pada integral adalah sebagai berikut :

(i)~\displaystyle{\int\limits {ax^n} \, dx=\frac{a}{n+1}x^{n+1}+C},~~~dengan~C=konstanta

(ii)~\displaystyle{\int\limits {kf(x)} \, dx=k\int\limits {f(x)} \, dx}

(iii)~\displaystyle{\int\limits {\left  f(x)\pm g(x) \right } \, dx=\int\limits {f(x)} \, dx\pm\int\limits {g(x)} \, dx}

(iv)~\displaystyle{\int\limits^b_a {f(x)} \, dx=F(b)-F(a)}

Untuk mengintegralkan fungsi rasional berbentuk \frac{f(x)}{g(x)} kita bisa dekomposisi fungsi tersebut terlebih dahulu menjadi pecahan pecahan parsial. Lalu kita integralkan tiap tiap pecahan parsialnya.

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DIKETAHUI

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

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DITANYA

Tentukan hasil integralnya.

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PENYELESAIAN

\frac{1}{x^4+1}=\frac{1}{x^4+2x^2-2x^2+1}

\frac{1}{x^4+1}=\frac{1}{x^4+2x^2+1-2x^2}

\frac{1}{x^4+1}=\frac{1}{(x^2+1)^2-(\sqrt{2}x)^2}

\frac{1}{x^4+1}=\frac{1}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}

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Misal :

\frac{1}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}=\frac{Ax+B}{x^2+\sqrt{2}x+1}+\frac{Cx+D}{x^2-\sqrt{2}x+1}

\frac{1}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}=\frac{(Ax+B)(x^2-\sqrt{2}x+1)+(Cx+D)(x^2+\sqrt{2}x+1)}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}

\frac{1}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}=\frac{(A+C)x^3+(-\sqrt{2}A+B+\sqrt{2}C+D)x^2+(A-\sqrt{2}B+C+\sqrt{2}D)x+(B+D)}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}

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Dengan menyamakan kedua ruas, kita peroleh 4 persamaan :

> Untuk koefisien x³ :

A+C=0

C=-A~~~~~~...(i)

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> Untuk konstanta :

B+D=1

D=1-B~~~~~~...(ii)

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> Untuk koefisien x² :

-\sqrt{2}A+B+\sqrt{2}C+D=0

-\sqrt{2}A+B+\sqrt{2}(\mathbf{-A})+(\mathbf{1-B})=0

-2\sqrt{2}A+1=0

A=\frac{1}{2\sqrt{2}}~\to~C=-\frac{1}{2\sqrt{2}}

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> Untuk koefsien x :

A-\sqrt{2}B+C+\sqrt{2}D=0

A-\sqrt{2}B+(\mathbf{-A})+\sqrt{2}(\mathbf{1-B})=0

-2\sqrt{2}B+\sqrt{2}=0

B=\frac{1}{2}~\to~D=\frac{1}{2}

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Diperoleh :

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

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

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Maka :

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

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

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

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

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

=\frac{1}{2\sqrt{2}}\left  \frac{1}{2}ln|2x^2+2\sqrt{2}x+2|+tan^{-1}(\sqrt{2}x+1)+C_3 \right +\frac{1}{2\sqrt{2}}-\frac{1}{2}ln|2x^2

.~~-2\sqrt{2}x+2|+tan^{-1}(\sqrt{2}x-1)+C_5

=\frac{1}{4\sqrt{2}}ln\left | \frac{2x^2+2\sqrt{2}x+2}{2x^2-2\sqrt{2}x+2} \right |+\frac{1}{2\sqrt{2}}\left  tan^{-1}(\sqrt{2}x+1)+tan^{-1}(\sqrt{2}x-1) \right +C

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KESIMPULAN

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

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PELAJARI LEBIH LANJUT

  1. Integral pecahan parsial : brainly.co.id/tugas/40289194
  2. Integral pecahan parsial : brainly.co.id/tugas/30067184
  3. Integral substitusi trigonometri : brainly.co.id/tugas/40357056

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DETAIL JAWABAN

Kelas : 11

Mapel: Matematika

Bab : Integral Tak Tentu

Kode Kategorisasi: 11.2.10

Kata Kunci : integral, antiturunan, fungsi, rasional, pecahan, parsial.


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