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

解

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

\frac{1}{8} ln ∣ 2 x + 1 ∣ - \frac{1}{8 ( 2 x + 1 )} - \frac{1}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{8 ( 2 x - 1 )} + C

解答ステップ

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

以下の部分分数を得る:

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

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

\int \frac{1}{4 ( 2 x + 1 )} d x = \frac{1}{8} ln ∣ 2 x + 1 ∣

\int \frac{1}{4 ( 2 x + 1 ) ^{2}} d x = - \frac{1}{8 ( 2 x + 1 )}

\int \frac{1}{4 ( 2 x - 1 )} d x = \frac{1}{8} ln ∣ 2 x - 1 ∣

\int \frac{1}{4 ( 2 x - 1 ) ^{2}} d x = - \frac{1}{8 ( 2 x - 1 )}

= \frac{1}{8} ln ∣ 2 x + 1 ∣ - \frac{1}{8 ( 2 x + 1 )} - \frac{1}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{8 ( 2 x - 1 )}

定数を解答に追加する

= \frac{1}{8} ln ∣ 2 x + 1 ∣ - \frac{1}{8 ( 2 x + 1 )} - \frac{1}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{8 ( 2 x - 1 )} + C