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

解

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

\frac{1}{3} arctan (\frac{x + 1}{3}) - \frac{1}{2 3} (arctan (\frac{1}{3} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{3} (x + 1)))) + C

解答ステップ

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

以下の部分分数を得る:

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

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

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

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

= \frac{1}{3} arctan (\frac{x + 1}{3}) - \frac{1}{2 3} (arctan (\frac{1}{3} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{3} (x + 1))))

定数を解答に追加する

= \frac{1}{3} arctan (\frac{x + 1}{3}) - \frac{1}{2 3} (arctan (\frac{1}{3} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{3} (x + 1)))) + C