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

解

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

- \frac{1}{16 x} + \frac{3}{128} ln ∣ x + 2 ∣ - \frac{1}{64 ( x + 2 )} - \frac{3}{128} ln ∣ x - 2 ∣ - \frac{1}{64 ( x - 2 )} + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{x ^{2} ( x ^{2} - 4 ) ^{2}} : \frac{1}{16 x ^{2}} + \frac{3}{128 ( x + 2 )} + \frac{1}{64 ( x + 2 ) ^{2}} - \frac{3}{128 ( x - 2 )} + \frac{1}{64 ( x - 2 ) ^{2}}

= \int \frac{1}{16 x ^{2}} + \frac{3}{128 ( x + 2 )} + \frac{1}{64 ( x + 2 ) ^{2}} - \frac{3}{128 ( x - 2 )} + \frac{1}{64 ( x - 2 ) ^{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}{16 x ^{2}} d x + \int \frac{3}{128 ( x + 2 )} d x + \int \frac{1}{64 ( x + 2 ) ^{2}} d x - \int \frac{3}{128 ( x - 2 )} d x + \int \frac{1}{64 ( x - 2 ) ^{2}} d x

\int \frac{1}{16 x ^{2}} d x = - \frac{1}{16 x}

\int \frac{3}{128 ( x + 2 )} d x = \frac{3}{128} ln ∣ x + 2 ∣

\int \frac{1}{64 ( x + 2 ) ^{2}} d x = - \frac{1}{64 ( x + 2 )}

\int \frac{3}{128 ( x - 2 )} d x = \frac{3}{128} ln ∣ x - 2 ∣

\int \frac{1}{64 ( x - 2 ) ^{2}} d x = - \frac{1}{64 ( x - 2 )}

= - \frac{1}{16 x} + \frac{3}{128} ln ∣ x + 2 ∣ - \frac{1}{64 ( x + 2 )} - \frac{3}{128} ln ∣ x - 2 ∣ - \frac{1}{64 ( x - 2 )}

定数を解答に追加する

= - \frac{1}{16 x} + \frac{3}{128} ln ∣ x + 2 ∣ - \frac{1}{64 ( x + 2 )} - \frac{3}{128} ln ∣ x - 2 ∣ - \frac{1}{64 ( x - 2 )} + C