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人気のある微分積分 >

integral of (16x^2)/((x-12)(x+4)^2)

解

\int \frac{16 x ^{2}}{( x - 12 ) ( x + 4 ) ^{2}} d x

解

9 ln ∣ x - 12 ∣ + 7 ln ∣ x + 4 ∣ + \frac{16}{x + 4} + C

解答ステップ

\int \frac{16 x ^{2}}{( x - 12 ) ( x + 4 ) ^{2}} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \int \frac{x ^{2}}{( x - 12 ) ( x + 4 ) ^{2}} d x

以下の部分分数を得る:

\frac{x ^{2}}{( x - 12 ) ( x + 4 ) ^{2}} : \frac{9}{16 ( x - 12 )} + \frac{7}{16 ( x + 4 )} - \frac{1}{( x + 4 ) ^{2}}

= 16 \cdot \int \frac{9}{16 ( x - 12 )} + \frac{7}{16 ( x + 4 )} - \frac{1}{( x + 4 ) ^{2}} d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 16 (\int \frac{9}{16 ( x - 12 )} d x + \int \frac{7}{16 ( x + 4 )} d x - \int \frac{1}{( x + 4 ) ^{2}} d x)

\int \frac{9}{16 ( x - 12 )} d x = \frac{9}{16} ln ∣ x - 12 ∣

\int \frac{7}{16 ( x + 4 )} d x = \frac{7}{16} ln ∣ x + 4 ∣

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

= 16 (\frac{9}{16} ln ∣ x - 12 ∣ + \frac{7}{16} ln ∣ x + 4 ∣ - (- \frac{1}{x + 4}))

簡素化

16 (\frac{9}{16} ln ∣ x - 12 ∣ + \frac{7}{16} ln ∣ x + 4 ∣ - (- \frac{1}{x + 4})) : 9 ln ∣ x - 12 ∣ + 7 ln ∣ x + 4 ∣ + \frac{16}{x + 4}

= 9 ln ∣ x - 12 ∣ + 7 ln ∣ x + 4 ∣ + \frac{16}{x + 4}

定数を解答に追加する

= 9 ln ∣ x - 12 ∣ + 7 ln ∣ x + 4 ∣ + \frac{16}{x + 4} + C

グラフ

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