integral of (e^{arctan(x)}-2x)/(1+x^2)

解

\int \frac{e ^{a r c t a n (x)} - 2 x}{1 + x ^{2}} d x

e^{a r c t a n (x)} - ln 1 + x^{2} + C

解答ステップ

\int \frac{e ^{a r c t a n (x)} - 2 x}{1 + x ^{2}} d x

拡張

\frac{e ^{a r c t a n (x)} - 2 x}{1 + x ^{2}} : \frac{e ^{a r c t a n (x)}}{1 + x ^{2}} - \frac{2 x}{1 + x ^{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{e ^{a r c t a n (x)}}{1 + x ^{2}} d x - \int \frac{2 x}{1 + x ^{2}} d x

\int \frac{e ^{a r c t a n (x)}}{1 + x ^{2}} d x = e^{a r c t a n (x)}

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

= e^{a r c t a n (x)} - ln 1 + x^{2}

定数を解答に追加する

= e^{a r c t a n (x)} - ln 1 + x^{2} + C