integral of 1/((4-x^2)^3)

解

\int \frac{1}{( 4 - x ^{2} ) ^{3}} d x

\frac{3}{512} ln ∣ x + 2 ∣ - \frac{3}{256 ( x + 2 )} - \frac{1}{128 ( x + 2 ) ^{2}} - \frac{3}{512} ln ∣ x - 2 ∣ - \frac{3}{256 ( x - 2 )} + \frac{1}{128 ( x - 2 ) ^{2}} + C

解答ステップ

\int \frac{1}{( 4 - x ^{2} ) ^{3}} d x

以下の部分分数を得る:

\frac{1}{( 4 - x ^{2} ) ^{3}} : \frac{3}{512 ( x + 2 )} + \frac{3}{256 ( x + 2 ) ^{2}} + \frac{1}{64 ( x + 2 ) ^{3}} - \frac{3}{512 ( x - 2 )} + \frac{3}{256 ( x - 2 ) ^{2}} - \frac{1}{64 ( x - 2 ) ^{3}}

= \int \frac{3}{512 ( x + 2 )} + \frac{3}{256 ( x + 2 ) ^{2}} + \frac{1}{64 ( x + 2 ) ^{3}} - \frac{3}{512 ( x - 2 )} + \frac{3}{256 ( x - 2 ) ^{2}} - \frac{1}{64 ( x - 2 ) ^{3}} 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{3}{512 ( x + 2 )} d x + \int \frac{3}{256 ( x + 2 ) ^{2}} d x + \int \frac{1}{64 ( x + 2 ) ^{3}} d x - \int \frac{3}{512 ( x - 2 )} d x + \int \frac{3}{256 ( x - 2 ) ^{2}} d x - \int \frac{1}{64 ( x - 2 ) ^{3}} d x

\int \frac{3}{512 ( x + 2 )} d x = \frac{3}{512} ln ∣ x + 2 ∣

\int \frac{3}{256 ( x + 2 ) ^{2}} d x = - \frac{3}{256 ( x + 2 )}

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

\int \frac{3}{512 ( x - 2 )} d x = \frac{3}{512} ln ∣ x - 2 ∣

\int \frac{3}{256 ( x - 2 ) ^{2}} d x = - \frac{3}{256 ( x - 2 )}

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

= \frac{3}{512} ln ∣ x + 2 ∣ - \frac{3}{256 ( x + 2 )} - \frac{1}{128 ( x + 2 ) ^{2}} - \frac{3}{512} ln ∣ x - 2 ∣ - \frac{3}{256 ( x - 2 )} - (- \frac{1}{128 ( x - 2 ) ^{2}})

簡素化

= \frac{3}{512} ln ∣ x + 2 ∣ - \frac{3}{256 ( x + 2 )} - \frac{1}{128 ( x + 2 ) ^{2}} - \frac{3}{512} ln ∣ x - 2 ∣ - \frac{3}{256 ( x - 2 )} + \frac{1}{128 ( x - 2 ) ^{2}}

定数を解答に追加する

= \frac{3}{512} ln ∣ x + 2 ∣ - \frac{3}{256 ( x + 2 )} - \frac{1}{128 ( x + 2 ) ^{2}} - \frac{3}{512} ln ∣ x - 2 ∣ - \frac{3}{256 ( x - 2 )} + \frac{1}{128 ( x - 2 ) ^{2}} + C