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人気のある微分積分 >

integral of (25x^2)/((x-3)(x+2)^2)

解

\int \frac{25 x ^{2}}{( x - 3 ) ( x + 2 ) ^{2}} d x

解

16 ln ∣ x + 2 ∣ + \frac{20}{x + 2} + 9 ln ∣ x - 3 ∣ + C

解答ステップ

\int \frac{25 x ^{2}}{( x - 3 ) ( x + 2 ) ^{2}} d x

定数を除く:

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

= 25 \cdot \int \frac{x ^{2}}{( x - 3 ) ( x + 2 ) ^{2}} d x

u置換積分法を適用する

= 25 \cdot \int \frac{( u - 2 ) ^{2}}{u ^{2} ( u - 5 )} d u

以下の部分分数を得る:

\frac{( u - 2 ) ^{2}}{u ^{2} ( u - 5 )} : \frac{16}{25 u} - \frac{4}{5 u ^{2}} + \frac{9}{25 ( u - 5 )}

= 25 \cdot \int \frac{16}{25 u} - \frac{4}{5 u ^{2}} + \frac{9}{25 ( u - 5 )} d u

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

= 25 (\int \frac{16}{25 u} d u - \int \frac{4}{5 u ^{2}} d u + \int \frac{9}{25 ( u - 5 )} d u)

\int \frac{16}{25 u} d u = \frac{16}{25} ln ∣ u ∣

\int \frac{4}{5 u ^{2}} d u = - \frac{4}{5 u}

\int \frac{9}{25 ( u - 5 )} d u = \frac{9}{25} ln ∣ u - 5 ∣

= 25 (\frac{16}{25} ln ∣ u ∣ - (- \frac{4}{5 u}) + \frac{9}{25} ln ∣ u - 5 ∣)

代用を戻す

u = x + 2

= 25 (\frac{16}{25} ln ∣ x + 2 ∣ - (- \frac{4}{5 ( x + 2 )}) + \frac{9}{25} ln ∣ x + 2 - 5 ∣)

簡素化

25 (\frac{16}{25} ln ∣ x + 2 ∣ - (- \frac{4}{5 ( x + 2 )}) + \frac{9}{25} ln ∣ x + 2 - 5 ∣) : 16 ln ∣ x + 2 ∣ + \frac{20}{x + 2} + 9 ln ∣ x - 3 ∣

= 16 ln ∣ x + 2 ∣ + \frac{20}{x + 2} + 9 ln ∣ x - 3 ∣

定数を解答に追加する

= 16 ln ∣ x + 2 ∣ + \frac{20}{x + 2} + 9 ln ∣ x - 3 ∣ + C

グラフ

人気の例

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