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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

簡素化

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

以下の部分分数を得る:

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

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

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

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

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

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

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

= (- \frac{1}{2}) (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )})

簡素化

(- \frac{1}{2}) (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )}) : - \frac{x ^{2} ln ∣ x - 1 ∣ - x ^{2} ln ∣ x + 1 ∣ - 2 x + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣}{8 ( x + 1 ) ( x - 1 )}

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

定数を解答に追加する

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

グラフ

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