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

解

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

\frac{1}{6} ln ∣ 4 x + 3 ∣ - \frac{2}{3} ln (4 ∣ x ∣) + C

解答ステップ

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

拡張

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

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

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

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

簡素化

2 (\frac{1}{3} ln ∣ 4 x + 3 ∣ - \frac{1}{3} ln (4 ∣ x ∣)) - \frac{1}{2} ln ∣ 4 x + 3 ∣ : \frac{1}{6} ln ∣ 4 x + 3 ∣ - \frac{2}{3} ln (4 ∣ x ∣)

= \frac{1}{6} ln ∣ 4 x + 3 ∣ - \frac{2}{3} ln (4 ∣ x ∣)

定数を解答に追加する

= \frac{1}{6} ln ∣ 4 x + 3 ∣ - \frac{2}{3} ln (4 ∣ x ∣) + C