Teknik Integral : Integral Parsial


Teknik pengintegralan parsial didasarkan pada pengintegralan rumus turunan hasil kali dua fungsi.

Teorema 3

Jika u = u(x) dan v = v(x) masing-masing fungsi terdiferensial, maka \int u dv = uv – \int v du.

Bukti :

Dari rumus turunan hasil kali dua fungsi diperoleh

\dfrac{d}{dx} (uv) = (\dfrac{d}{dx} u(x)) v(x) + (\dfrac{d}{dx} v(x)) u(x)

= v \dfrac{du}{dx} + u \dfrac{dv}{dx}

Pengintegralan pada kedua ruas menghasilkan

\int (\dfrac{d}{dx} (uv))dx = \int (v \dfrac{du}{dx}) + \int (u \dfrac{dv}{dx})

\Leftrightarrow uv = \int v du + \int u dv

\Leftrightarrow u dv = uv + \int v du \blacksquare

Contoh :

  1. \int xe-x dx = …

    ambil u = x dan dv = e-x dx \Leftrightarrow v = -e-x, maka

    \int xe-x dx = \int u dv = uv – \int v du

    = -xe-x\int e-x dx

    = -xe-x – e-x + C

    Atau bisa juga dikerjakan seperti ini

    \int xe-x dx = \int x d\dfrac{e^{-x}}{-e^{-x}}

    = –\int x d(e-x)

    = -(xe-x\int e-x dx)

    = -(xe-x + e-x) + C

    = -xe-x + e-x + C

  2. \int ex sin x dx = …

    \int ex sin x dx = \int sin x d(ex)

    = ex sin x – \int ex d(sin x)

    = ex sin x – \int ex cos x dx

    = ex sin x – \int cos x d(ex)

    = ex sin x – [ex cos x – \int ex d(cos x)]

    = ex sin x – [ex cos x – \int – ex sin x]

    = ex sin x – ex cos x – \int ex sin x dx

    sehingga diperoleh,

    = 2 \int ex sin x dx

    = ex sin x – ex cos x

    \Leftrightarrow \int ex sin x dx = \dfrac{1}{2}(ex sin x – ex cos x) + C

  3. \int ln x dx = …

    \int ln x dx = x ln x – \int x d(ln x)

    = x ln x – \int x \dfrac{1}{x} dx

    = x ln x – \int dx

    = x ln x – x + C

  4. \int x \sqrt{x+3} dx = …

    $latex \int$ x \sqrt{x+3} dx = \int x \dfrac{2}{3} d(x + 3)3/2

    = \dfrac{2}{3} \int x d(x + 3)3/2

    = \dfrac{2}{3} [x(x + 3)3/2\int (x + 3)3/2 dx]

    = \dfrac{2}{3} [x(x + 3)3/2\int (x + 3)3/2 d(x + 3)]

    = \dfrac{2}{3} [x(x + 3)3/2\dfrac{2}{5} (x + 3)3/2] + C

2 comments on “Teknik Integral : Integral Parsial

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