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

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

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

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

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

定数を解答に追加する

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