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人気のある微分積分 >

integral of 1/((x+4)^2(x-1))

解

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

解

- \frac{1}{25} ln ∣ x + 4 ∣ + \frac{1}{5 ( x + 4 )} + \frac{1}{25} ln ∣ x - 1 ∣ + C

解答ステップ

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

u置換積分法を適用する

= \int \frac{1}{u ^{2} ( u - 5 )} d u

以下の部分分数を得る:

\frac{1}{u ^{2} ( u - 5 )} : - \frac{1}{25 u} - \frac{1}{5 u ^{2}} + \frac{1}{25 ( u - 5 )}

= \int - \frac{1}{25 u} - \frac{1}{5 u ^{2}} + \frac{1}{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

= - \int \frac{1}{25 u} d u - \int \frac{1}{5 u ^{2}} d u + \int \frac{1}{25 ( u - 5 )} d u

\int \frac{1}{25 u} d u = \frac{1}{25} ln ∣ u ∣

\int \frac{1}{5 u ^{2}} d u = - \frac{1}{5 u}

\int \frac{1}{25 ( u - 5 )} d u = \frac{1}{25} ln ∣ u - 5 ∣

= - \frac{1}{25} ln ∣ u ∣ - (- \frac{1}{5 u}) + \frac{1}{25} ln ∣ u - 5 ∣

代用を戻す

u = x + 4

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

簡素化

- \frac{1}{25} ln ∣ x + 4 ∣ - (- \frac{1}{5 ( x + 4 )}) + \frac{1}{25} ln ∣ x + 4 - 5 ∣ : - \frac{1}{25} ln ∣ x + 4 ∣ + \frac{1}{5 ( x + 4 )} + \frac{1}{25} ln ∣ x - 1 ∣

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

定数を解答に追加する

= - \frac{1}{25} ln ∣ x + 4 ∣ + \frac{1}{5 ( x + 4 )} + \frac{1}{25} ln ∣ x - 1 ∣ + C

グラフ

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