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

解

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

\frac{1}{2} ln x^{2} + 1 - arctan (x) + C

解答ステップ

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

拡張

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

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

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

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

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

定数を解答に追加する

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