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人気のある微分積分 >

integral of (180x^2)/((3x+1)(3x+5)(x+2))

解

\int \frac{180 x ^{2}}{( 3 x + 1 ) ( 3 x + 5 ) ( x + 2 )} d x

解

ln ∣ 3 x + 1 ∣ - 125 ln ∣ 3 x + 5 ∣ + 144 ln ∣ x + 2 ∣ + C

解答ステップ

\int \frac{180 x ^{2}}{( 3 x + 1 ) ( 3 x + 5 ) ( x + 2 )} d x

定数を除く:

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

= 180 \cdot \int \frac{x ^{2}}{( 3 x + 1 ) ( 3 x + 5 ) ( x + 2 )} d x

以下の部分分数を得る:

\frac{x ^{2}}{( 3 x + 1 ) ( 3 x + 5 ) ( x + 2 )} : \frac{1}{60 ( 3 x + 1 )} - \frac{25}{12 ( 3 x + 5 )} + \frac{4}{5 ( x + 2 )}

= 180 \cdot \int \frac{1}{60 ( 3 x + 1 )} - \frac{25}{12 ( 3 x + 5 )} + \frac{4}{5 ( x + 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

= 180 (\int \frac{1}{60 ( 3 x + 1 )} d x - \int \frac{25}{12 ( 3 x + 5 )} d x + \int \frac{4}{5 ( x + 2 )} d x)

\int \frac{1}{60 ( 3 x + 1 )} d x = \frac{1}{180} ln ∣ 3 x + 1 ∣

\int \frac{25}{12 ( 3 x + 5 )} d x = \frac{25}{36} ln ∣ 3 x + 5 ∣

\int \frac{4}{5 ( x + 2 )} d x = \frac{4}{5} ln ∣ x + 2 ∣

= 180 (\frac{1}{180} ln ∣ 3 x + 1 ∣ - \frac{25}{36} ln ∣ 3 x + 5 ∣ + \frac{4}{5} ln ∣ x + 2 ∣)

簡素化

180 (\frac{1}{180} ln ∣ 3 x + 1 ∣ - \frac{25}{36} ln ∣ 3 x + 5 ∣ + \frac{4}{5} ln ∣ x + 2 ∣) : ln ∣ 3 x + 1 ∣ - 125 ln ∣ 3 x + 5 ∣ + 144 ln ∣ x + 2 ∣

= ln ∣ 3 x + 1 ∣ - 125 ln ∣ 3 x + 5 ∣ + 144 ln ∣ x + 2 ∣

定数を解答に追加する

= ln ∣ 3 x + 1 ∣ - 125 ln ∣ 3 x + 5 ∣ + 144 ln ∣ x + 2 ∣ + C

グラフ

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