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人気のある微分積分 >

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

解

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

解

- \frac{12}{125} ln ∣ x + 2 ∣ - \frac{4}{25 ( x + 2 )} + \frac{12}{125} ln ∣ x - 3 ∣ - \frac{9}{25 ( x - 3 )} + C

解答ステップ

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

u置換積分法を適用する

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

以下の部分分数を得る:

\frac{( u - 2 ) ^{2}}{u ^{2} ( u - 5 ) ^{2}} : - \frac{12}{125 u} + \frac{4}{25 u ^{2}} + \frac{12}{125 ( u - 5 )} + \frac{9}{25 ( u - 5 ) ^{2}}

= \int - \frac{12}{125 u} + \frac{4}{25 u ^{2}} + \frac{12}{125 ( u - 5 )} + \frac{9}{25 ( u - 5 ) ^{2}} 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

= - \int \frac{12}{125 u} d u + \int \frac{4}{25 u ^{2}} d u + \int \frac{12}{125 ( u - 5 )} d u + \int \frac{9}{25 ( u - 5 ) ^{2}} d u

\int \frac{12}{125 u} d u = \frac{12}{125} ln ∣ u ∣

\int \frac{4}{25 u ^{2}} d u = - \frac{4}{25 u}

\int \frac{12}{125 ( u - 5 )} d u = \frac{12}{125} ln ∣ u - 5 ∣

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

= - \frac{12}{125} ln ∣ u ∣ - \frac{4}{25 u} + \frac{12}{125} ln ∣ u - 5 ∣ - \frac{9}{25 ( u - 5 )}

代用を戻す

u = x + 2

= - \frac{12}{125} ln ∣ x + 2 ∣ - \frac{4}{25 ( x + 2 )} + \frac{12}{125} ln ∣ x + 2 - 5 ∣ - \frac{9}{25 ( x + 2 - 5 )}

簡素化

- \frac{12}{125} ln ∣ x + 2 ∣ - \frac{4}{25 ( x + 2 )} + \frac{12}{125} ln ∣ x + 2 - 5 ∣ - \frac{9}{25 ( x + 2 - 5 )} : - \frac{12}{125} ln ∣ x + 2 ∣ - \frac{4}{25 ( x + 2 )} + \frac{12}{125} ln ∣ x - 3 ∣ - \frac{9}{25 ( x - 3 )}

= - \frac{12}{125} ln ∣ x + 2 ∣ - \frac{4}{25 ( x + 2 )} + \frac{12}{125} ln ∣ x - 3 ∣ - \frac{9}{25 ( x - 3 )}

定数を解答に追加する

= - \frac{12}{125} ln ∣ x + 2 ∣ - \frac{4}{25 ( x + 2 )} + \frac{12}{125} ln ∣ x - 3 ∣ - \frac{9}{25 ( x - 3 )} + C

グラフ

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