integral of (4x^3-x^2+4x+2)/((x^2+1)^2)

解

\int \frac{4 x ^{3} - x ^{2} + 4 x + 2}{( x ^{2} + 1 ) ^{2}} d x

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

解答ステップ

\int \frac{4 x ^{3} - x ^{2} + 4 x + 2}{( x ^{2} + 1 ) ^{2}} d x

拡張

\frac{4 x ^{3} - x ^{2} + 4 x + 2}{( x ^{2} + 1 ) ^{2}} : \frac{4 x ^{3}}{( x ^{2} + 1 ) ^{2}} - \frac{x ^{2}}{( x ^{2} + 1 ) ^{2}} + \frac{4 x}{( x ^{2} + 1 ) ^{2}} + \frac{2}{( 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 ^{3}}{( x ^{2} + 1 ) ^{2}} d x - \int \frac{x ^{2}}{( x ^{2} + 1 ) ^{2}} d x + \int \frac{4 x}{( x ^{2} + 1 ) ^{2}} d x + \int \frac{2}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{4 x ^{3}}{( x ^{2} + 1 ) ^{2}} d x = 2 (ln x^{2} + 1 + \frac{1}{x ^{2} + 1})

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

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

簡素化

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

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

定数を解答に追加する

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