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

解

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

x - 4 ln ∣ x + 1 ∣ + 2 ln ∣ x ∣ + \frac{1}{x} + C

解答ステップ

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

拡張

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

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

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

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

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

簡素化

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

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

定数を解答に追加する

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