-->

Contoh Soal dan Pembahasan Integral Fungsi Pecah

Topik Bahasan
Tentang bagaimana cara memecah fungsi, anda sebaiknya membaca: di:  Fungsi Pecah. Berikut untuk integral fungsi pecah juga kembali harus diingat,
Rumus dasar integral dan sifat logartima natural (ln) yang harus anda kembali ingat,
 $ \int k(ax+b)^n dx = \frac{k}{a} . \frac{1}{n+1} (ax+b)^{n+1} + c $
 $ \int \frac{k}{ax+b} dx = \frac{k}{a} . \ln (ax+b)+ c $
Juga anda ingat kembali sifat logarima yang diterapkan pada logaritma natural (ln).
$ \ln a + \ln b = \ln (a.b) $
$ \ln a - \ln b = \ln \frac{a}{b} $

Contoh 1: $ \int \frac{2x + 1}{x^2 - 3x} dx = ...$
$ \int \frac{2x + 1}{x^2 - 3x} dx = \int \frac{1}{6} \left( \frac{-2}{x} + \frac{21}{x-3} \right) dx $
Pemecahan Fungsi
Faktorkan Penyebut:
$ x^2 - 3x = x(x-3) $.
Bagi menjadi dua bagian,
$ \begin{align} \frac{2x + 1}{x^2 - 3x} & = \frac{2x + 1}{x(x-3)} = \frac{A}{x} + \frac{B}{x-3} \\ & = \frac{A(x-3) + Bx}{x(x-3)} \\ & = \frac{Ax - 3A + Bx}{x(x-3)} \\ \frac{2x + 1}{x^2 - 3x} & = \frac{(A+B)x - 3A }{x(x-3)} \\ 2x + 1 & = (A+B)x - 3A \end{align} $
Tentukan nilai A dan B melalui kesamaan
 $ 2x + 1 = (A+B)x - 3A $,
$ -3A = 1 \rightarrow A = -\frac{1}{3} $
$ A + B = 2 \rightarrow -\frac{1}{3} + B = 2 \rightarrow B = \frac{7}{2} $
Jadi bentuk pemecahannya,
$ \begin{align} \frac{2x + 1}{x^2 - 3x} & = \frac{A}{x} + \frac{B}{x-3} \\ \frac{2x + 1}{x^2 - 3x} & = \frac{-\frac{1}{3}}{x} + \frac{\frac{7}{2}}{x-3} \\ \frac{2x + 1}{x^2 - 3x} & = \frac{1}{6} \left( \frac{-2}{x} + \frac{21}{x-3} \right) \end{align} $
Bisa ditulis,
 $ \begin{align} \frac{2x + 1}{x^2 - 3x} & = \frac{1}{6} \left( \frac{-2}{x} + \frac{21}{x-3} \right) \end{align} $

Integralkan:
$ \begin{align} \int \frac{2x + 1}{x^2 - 3x} dx & = \int \frac{1}{6} \left( \frac{-2}{x} + \frac{21}{x-3} \right) dx \\ & = \frac{1}{6} \left( \int \frac{-2}{x} dx + \int \frac{21}{x-3} dx \right) \\ & = \frac{1}{6} \left( -2 \ln x + 21 \ln (x-3) \right) + c \\ & = \frac{1}{6} \left( \ln x^{-2} + \ln (x-3)^{21} \right) + c \\ & = \frac{1}{6} \left( \ln \frac{1}{x^2} + \ln (x-3)^{21} \right) + c \\ & = \frac{1}{6} \left( \ln (\frac{1}{x^2} (x-3)^{21} ) \right) + c \\ & = \frac{1}{6} \ln \left( \frac{(x-3)^{21}}{x^2} \right) + c \end{align} $
Jadi,$ \begin{align} \int \frac{2x + 1}{x^2 - 3x} dx & = \frac{1}{6} \ln \left( \frac{(x-3)^{21}}{x^2} \right) + c \end{align} $

Contoh 2: Tentukan hasil dari integral $ \int \frac{3x^2 - x }{(x+1)(x-1)^2 } dx $
Penyelesaian :
Lakukan pemecahan fungsi:
$ \begin{align} \frac{3x^2 - x }{(x+1)(x-1)^2 } & = \frac{A}{x+1} + \frac{B }{x-1} + \frac{C}{(x-1)^2} \\ & = \frac{A(x-1)^2 + B(x+1)(x-1) + C(x+1)}{(x+1)(x-1)^2} \\ & = \frac{A(x^2 - 2x + 1) + B(x^2 - 1) + C(x+1)}{(x+1)(x-1)^2} \\ & = \frac{ Ax^2 - 2Ax + A + Bx^2 - B + Cx+C}{(x+1)(x-1)^2} \\ \frac{3x^2 - x }{(x+1)(x-1)^2 } & = \frac{ (A+B)x^2 +(C-2A)x + (A-B+c) }{(x+1)(x-1)^2} \\ 3x^2 - x & = (A+B)x^2 +(C-2A)x + (A-B+c) \end{align} $
Tentukan nilai A,B dan C dengan kesamaan,
 $ 3x^2 - x = (A+B)x^2 +(C-2A)x + (A-B+c) $,
$ A + B = 3 \rightarrow B = 3 - A \, $ ....pers(i)
$ C - 2A = -1 \rightarrow C = 2A - 1 \, $ ....pers(ii)
$ A - B + C = 0 \, $ ....pers(iii)
Masukkan pers(i) dan (ii) ke pers(iii)
$ \begin{align} A - B + C & = 0 \\ A - (3 - A) + (2A - 1) & = 0 \\ 4A - 4 & = 0 \\ A & = 1 \end{align} $
Pers(i) : $ B = 3 - A = 3 - 1 = 2 $
Pers(ii) : $ C = 2A - 1 = 2.1 - 1 = 1 $
Ditemukan bentuk pecahan :
$ \begin{align} \frac{3x^2 - x }{(x+1)(x-1)^2 } & = \frac{A}{x+1} + \frac{B }{x-1} + \frac{C}{(x-1)^2} \\ \frac{3x^2 - x }{(x+1)(x-1)^2 } & = \frac{1}{x+1} + \frac{2 }{x-1} + \frac{1}{(x-1)^2} \end{align} $
Lanjutkan dengan mengintegralkan:
$ \begin{align} \int \frac{3x^2 - x }{(x+1)(x-1)^2 } dx & = \int \frac{1}{x+1} + \frac{2 }{x-1} + \frac{1}{(x-1)^2} dx \\ & = \int \frac{1}{x+1} dx + \int \frac{2 }{x-1} dx + \int \frac{1}{(x-1)^2} dx \\ & = \ln (x+1) + 2 \ln (x-1) + \int (x-1)^{-2} dx \\ & = \ln (x+1) + \ln (x-1)^2 + \frac{1}{-2+1} (x-1)^{-2+1} + c \\ & = \ln [(x+1) (x-1)^2 ] + \frac{1}{-1} (x-1)^{-1} + c \\ & = \ln [(x+1) (x-1)^2 ] -  \frac{1}{(x-1)} + c \end{align} $
Jadi, $ \begin{align} \int \frac{3x^2 - x }{(x+1)(x-1)^2 } dx & = \ln [(x+1) (x-1)^2 ] - \frac{1}{(x-1)} + c \end{align} $
.

Cari Soal dan Pembahasan tentang

Loading...