integral of (4x^2+4x+4)/((x^2+1)^2)

解

\int \frac{4 x ^{2} + 4 x + 4}{( x ^{2} + 1 ) ^{2}} d x

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

解答ステップ

\int \frac{4 x ^{2} + 4 x + 4}{( x ^{2} + 1 ) ^{2}} d x

拡張

\frac{4 x ^{2} + 4 x + 4}{( x ^{2} + 1 ) ^{2}} : \frac{4 x ^{2}}{( x ^{2} + 1 ) ^{2}} + \frac{4 x}{( x ^{2} + 1 ) ^{2}} + \frac{4}{( x ^{2} + 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{4 x ^{2}}{( x ^{2} + 1 ) ^{2}} d x + \int \frac{4 x}{( x ^{2} + 1 ) ^{2}} d x + \int \frac{4}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{4 x ^{2}}{( x ^{2} + 1 ) ^{2}} d x = 4 (\frac{1}{2} arctan (x) - \frac{x}{2 ( 1 + x ^{2} )})

\int \frac{4 x}{( x ^{2} + 1 ) ^{2}} d x = - \frac{2}{x ^{2} + 1}

\int \frac{4}{( x ^{2} + 1 ) ^{2}} d x = 2 (arctan (x) + \frac{x}{1 + x ^{2}})

= 4 (\frac{1}{2} arctan (x) - \frac{x}{2 ( 1 + x ^{2} )}) - \frac{2}{x ^{2} + 1} + 2 (arctan (x) + \frac{x}{1 + x ^{2}})

簡素化

4 (\frac{1}{2} arctan (x) - \frac{x}{2 ( 1 + x ^{2} )}) - \frac{2}{x ^{2} + 1} + 2 (arctan (x) + \frac{x}{1 + x ^{2}}) : 4 arctan (x) - \frac{2}{x ^{2} + 1}

= 4 arctan (x) - \frac{2}{x ^{2} + 1}

定数を解答に追加する

= 4 arctan (x) - \frac{2}{x ^{2} + 1} + C