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

解

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

\frac{5}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x)) + C

解答ステップ

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

拡張

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

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

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

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

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

簡素化

\frac{1}{2} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))) + 2 arcsin (x) : \frac{5}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x))

= \frac{5}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x))

定数を解答に追加する

= \frac{5}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x)) + C