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

解

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

ln x^{2} - 2 x + 2 + C

解答ステップ

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

拡張

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

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

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

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

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

簡素化

= ln x^{2} - 2 x + 2

定数を解答に追加する

= ln x^{2} - 2 x + 2 + C