integral of (1+x)/(sqrt(9-4x^2))

解

\int \frac{1 + x}{9 - 4 x ^{2}} d x

\frac{1}{2} arcsin (\frac{2}{3} x) - \frac{1}{4} 9 - 4 x^{2} + C

解答ステップ

\int \frac{1 + x}{9 - 4 x ^{2}} d x

拡張

\frac{1 + x}{9 - 4 x ^{2}} : \frac{1}{9 - 4 x ^{2}} + \frac{x}{9 - 4 x ^{2}}

= \int \frac{1}{9 - 4 x ^{2}} + \frac{x}{9 - 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}{9 - 4 x ^{2}} d x + \int \frac{x}{9 - 4 x ^{2}} d x

\int \frac{1}{9 - 4 x ^{2}} d x = \frac{1}{2} arcsin (\frac{2}{3} x)

\int \frac{x}{9 - 4 x ^{2}} d x = - \frac{1}{4} 9 - 4 x^{2}

= \frac{1}{2} arcsin (\frac{2}{3} x) - \frac{1}{4} 9 - 4 x^{2}

定数を解答に追加する

= \frac{1}{2} arcsin (\frac{2}{3} x) - \frac{1}{4} 9 - 4 x^{2} + C