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

解

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

x + \frac{x ^{2}}{2} + C

解答ステップ

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

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

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

簡素化

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

= \int 1 + x 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 1 d x + \int x d x

\int 1 d x = x

\int x d x = \frac{x ^{2}}{2}

= x + \frac{x ^{2}}{2}

定数を解答に追加する

= x + \frac{x ^{2}}{2} + C