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hitung hasil dari[tex]∫ \frac{x + 3}{ \sqrt{3 + 4x -

Berikut ini adalah pertanyaan dari irsyadnurrohmapcz6u3 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Hitung hasil dari
∫ \frac{x + 3}{ \sqrt{3 + 4x - 9 {x}^{2} } } dx
​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Hasil dari \displaystyle{\int\limits {\frac{x+3}{\sqrt{3+4x-9x^2}}} \, dx }adalah\displaystyle{\boldsymbol{\frac{1}{27}\left [ -3\sqrt{3+4x-9x^2}+29arcsin\left ( \frac{9x-2}{\sqrt{31}} \right ) \right ]+C} }.

PEMBAHASAN

Substitusi trigonometri dapat digunakan untuk menyelesaikan integral dengan bentuk :

\displaystyle{\sqrt{a^2-x^2}~\to~substitusi~sin\theta=\frac{x}{a}}

\displaystyle{\sqrt{x^2-a^2}~\to~substitusi~sec\theta=\frac{x}{a}}

\displaystyle{\sqrt{x^2+a^2}~\to~substitusi~tan\theta=\frac{x}{a}}

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DIKETAHUI

\displaystyle{\int\limits {\frac{x+3}{\sqrt{3+4x-9x^2}}} \, dx= }

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DITANYA

Tentukan hasilnya.

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PENYELESAIAN

\displaystyle{\int\limits {\frac{x+3}{\sqrt{3+4x-9x^2}}} \, dx }

\displaystyle{=\int\limits {\frac{x+3}{\sqrt{-9\left ( x^2-\frac{4}{9}x-\frac{1}{3} \right )}}} \, dx }

\displaystyle{=\int\limits {\frac{x+3}{\sqrt{-9\left ( x^2-\frac{4}{9}x+\frac{4}{81}-\frac{1}{3}-\frac{4}{81} \right )}}} \, dx }

\displaystyle{=\int\limits {\frac{x+3}{\sqrt{-9\left [ \left ( x-\frac{2}{9} \right )^2-\frac{31}{81} \right ]}}} \, dx }

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Misal :

\displaystyle{u=x-\frac{2}{9}~\to~du=dx}

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\displaystyle{=\int\limits {\frac{u+\frac{2}{9}+3}{\sqrt{-9\left ( u^2-\frac{31}{81} \right )}}} \, du }

\displaystyle{=\int\limits {\frac{u+\frac{29}{9}}{\sqrt{9\left [ \left ( \frac{\sqrt{31}}{9} \right )^2-u^2 \right ]}}} \, du }

\displaystyle{=\int\limits {\frac{u+\frac{29}{9}}{3\sqrt{\left ( \frac{\sqrt{31}}{9} \right )^2-u^2}}} \, du }

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Misal :

\displaystyle{sin\theta=\frac{u}{\frac{\sqrt{31}}{9}} }

\displaystyle{sin\theta=\frac{9u}{\sqrt{31}} }

\displaystyle{cos\theta d\theta=\frac{9}{\sqrt{31}}du }

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\displaystyle{=\int\limits {\frac{\frac{\sqrt{31}}{9}sin\theta+\frac{29}{9}}{3\sqrt{\left ( \frac{\sqrt{31}}{9} \right )^2-\left ( \frac{\sqrt{31}}{9}sin\theta \right )^2}}} \, \left ( \frac{\sqrt{31}}{9}cos\theta d\theta \right ) }

\displaystyle{=\frac{1}{3}\times\frac{\sqrt{31}}{9}\int\limits {\frac{\frac{1}{9}\left ( \sqrt{31}sin\theta+29 \right )cos\theta}{\sqrt{\frac{31}{81}-\frac{31}{81}sin^2\theta}}} \, d\theta }

\displaystyle{=\frac{\sqrt{31}}{27}\times\frac{1}{9}\int\limits {\frac{\left ( \sqrt{31}sin\theta+29 \right )cos\theta}{\sqrt{\frac{31}{81}\left ( 1-sin^2\theta \right )}}} \, d\theta }

\displaystyle{=\frac{\sqrt{31}}{243}\int\limits {\frac{\left ( \sqrt{31}sin\theta+29 \right )cos\theta}{\frac{\sqrt{31}}{9}cos\theta}} \, d\theta }

\displaystyle{=\frac{\sqrt{31}}{243}\times\frac{9}{\sqrt{31}}\int\limits {\left ( \sqrt{31}sin\theta+29 \right )} \, d\theta }

\displaystyle{=\frac{1}{27}\int\limits {\left ( \sqrt{31}sin\theta+29 \right )} \, d\theta }

\displaystyle{=\frac{1}{27}\left ( -\sqrt{31}cos\theta+29\theta \right )+C }

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Mencari cos\theta :

\displaystyle{sin\theta=\frac{9u}{\sqrt{31}} }

\displaystyle{\frac{sisi~depan}{sisi~miring}=\frac{9u}{\sqrt{31}} }

sisi~samping=\sqrt{(sisi~miring)^2-(sisi~depan)^2}

sisi~samping=\sqrt{(\sqrt{31})^2-(9u)^2}

sisi~samping=\sqrt{31-81u^2}

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Maka :

\displaystyle{cos\theta=\frac{sisi~samping}{sisi~miring}}

\displaystyle{cos\theta=\frac{\sqrt{31-81u^2}}{\sqrt{31}}}

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\displaystyle{=\frac{1}{27}\left [ -\sqrt{31}\left ( \frac{\sqrt{31-81u^2}}{\sqrt{31}} \right )+29arcsin\left ( \frac{9u}{\sqrt{31}} \right ) \right ]+C }

\displaystyle{=\frac{1}{27}\left [ -\sqrt{31-81u^2}+29arcsin\left ( \frac{9u}{\sqrt{31}} \right ) \right ]+C }

\displaystyle{=\frac{1}{27}\left [ -\sqrt{31-81\left ( x-\frac{2}{9} \right )^2}+29arcsin\left ( \frac{9\left ( x-\frac{2}{9} \right )}{\sqrt{31}} \right ) \right ]+C }

\displaystyle{=\frac{1}{27}\left [ -\sqrt{31-81\left ( x^2-\frac{4}{9}x+\frac{4}{81} \right )}+29arcsin\left ( \frac{9x-2}{\sqrt{31}} \right ) \right ]+C }

\displaystyle{=\frac{1}{27}\left [ -\sqrt{31-81x^2+36x-4}+29arcsin\left ( \frac{9x-2}{\sqrt{31}} \right ) \right ]+C }

\displaystyle{=\frac{1}{27}\left [ -\sqrt{27+36x-81x^2}+29arcsin\left ( \frac{9x-2}{\sqrt{31}} \right ) \right ]+C }

\displaystyle{=\frac{1}{27}\left [ -\sqrt{9(3+4x-9x^2)}+29arcsin\left ( \frac{9x-2}{\sqrt{31}} \right ) \right ]+C }

\displaystyle{=\frac{1}{27}\left [ -3\sqrt{3+4x-9x^2}+29arcsin\left ( \frac{9x-2}{\sqrt{31}} \right ) \right ]+C }

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PELAJARI LEBIH LANJUT

  1. Integral substitusi : yomemimo.com/tugas/40327197
  2. Integral substitusi : yomemimo.com/tugas/30251199

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DETAIL JAWABAN

Kelas : 11

Mapel: Matematika

Bab : Integral

Kode Kategorisasi: 11.2.10.

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Last Update: Sun, 24 Jul 22
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