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

解

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

\frac{1}{4} x^{2} + \frac{1}{4} x - \frac{3}{8} ln ∣ - 2 x + 1 ∣ - \frac{3}{16} + C

解答ステップ

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

拡張

\frac{1 - x ^{2}}{1 - 2 x} : \frac{1}{1 - 2 x} - \frac{x ^{2}}{1 - 2 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}{1 - 2 x} d x - \int \frac{x ^{2}}{1 - 2 x} d x

\int \frac{1}{1 - 2 x} d x = - \frac{1}{2} ln ∣ 1 - 2 x ∣

\int \frac{x ^{2}}{1 - 2 x} d x = - \frac{1}{16} (4 x^{2} + 4 x + 2 ln ∣ - 2 x + 1 ∣ - 3)

= - \frac{1}{2} ln ∣ 1 - 2 x ∣ - (- \frac{1}{16} (4 x^{2} + 4 x + 2 ln ∣ - 2 x + 1 ∣ - 3))

簡素化

- \frac{1}{2} ln ∣ 1 - 2 x ∣ - (- \frac{1}{16} (4 x^{2} + 4 x + 2 ln ∣ - 2 x + 1 ∣ - 3)) : \frac{1}{4} x^{2} + \frac{1}{4} x - \frac{3}{8} ln ∣ - 2 x + 1 ∣ - \frac{3}{16}

= \frac{1}{4} x^{2} + \frac{1}{4} x - \frac{3}{8} ln ∣ - 2 x + 1 ∣ - \frac{3}{16}

定数を解答に追加する

= \frac{1}{4} x^{2} + \frac{1}{4} x - \frac{3}{8} ln ∣ - 2 x + 1 ∣ - \frac{3}{16} + C