integral of 1/((x^2-25)^2)

解

\int \frac{1}{( x ^{2} - 25 ) ^{2}} d x

\frac{1}{500} ln ∣ x + 5 ∣ - \frac{1}{100 ( x + 5 )} - \frac{1}{500} ln ∣ x - 5 ∣ - \frac{1}{100 ( x - 5 )} + C

解答ステップ

\int \frac{1}{( x ^{2} - 25 ) ^{2}} d x

以下の部分分数を得る:

\frac{1}{( x ^{2} - 25 ) ^{2}} : \frac{1}{500 ( x + 5 )} + \frac{1}{100 ( x + 5 ) ^{2}} - \frac{1}{500 ( x - 5 )} + \frac{1}{100 ( x - 5 ) ^{2}}

= \int \frac{1}{500 ( x + 5 )} + \frac{1}{100 ( x + 5 ) ^{2}} - \frac{1}{500 ( x - 5 )} + \frac{1}{100 ( x - 5 ) ^{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

= \int \frac{1}{500 ( x + 5 )} d x + \int \frac{1}{100 ( x + 5 ) ^{2}} d x - \int \frac{1}{500 ( x - 5 )} d x + \int \frac{1}{100 ( x - 5 ) ^{2}} d x

\int \frac{1}{500 ( x + 5 )} d x = \frac{1}{500} ln ∣ x + 5 ∣

\int \frac{1}{100 ( x + 5 ) ^{2}} d x = - \frac{1}{100 ( x + 5 )}

\int \frac{1}{500 ( x - 5 )} d x = \frac{1}{500} ln ∣ x - 5 ∣

\int \frac{1}{100 ( x - 5 ) ^{2}} d x = - \frac{1}{100 ( x - 5 )}

= \frac{1}{500} ln ∣ x + 5 ∣ - \frac{1}{100 ( x + 5 )} - \frac{1}{500} ln ∣ x - 5 ∣ - \frac{1}{100 ( x - 5 )}

定数を解答に追加する

= \frac{1}{500} ln ∣ x + 5 ∣ - \frac{1}{100 ( x + 5 )} - \frac{1}{500} ln ∣ x - 5 ∣ - \frac{1}{100 ( x - 5 )} + C