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

解

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

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

解答ステップ

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

拡張

\frac{1 - 2 x}{1 - 4 x ^{2}} : \frac{1}{1 - 4 x ^{2}} - \frac{2 x}{1 - 4 x ^{2}}

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

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

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

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

簡素化

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

定数を解答に追加する

= \frac{1}{2} arcsin (2 x) + \frac{1}{2} 1 - 4 x^{2} + C