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

解

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

\frac{5}{2} arctan (x) + \frac{x}{2 ( 1 + x ^{2} )} + C

解答ステップ

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

以下の部分分数を得る:

\frac{2 x ^{2} + 3}{( x ^{2} + 1 ) ^{2}} : \frac{2}{x ^{2} + 1} + \frac{1}{( x ^{2} + 1 ) ^{2}}

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

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

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

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

簡素化

2 arctan (x) + \frac{1}{2} (arctan (x) + \frac{x}{1 + x ^{2}}) : \frac{5}{2} arctan (x) + \frac{x}{2 ( 1 + x ^{2} )}

= \frac{5}{2} arctan (x) + \frac{x}{2 ( 1 + x ^{2} )}

定数を解答に追加する

= \frac{5}{2} arctan (x) + \frac{x}{2 ( 1 + x ^{2} )} + C