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

解

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

\frac{1}{2} ln x^{2} + 2 x + 2 - \frac{5}{2} arctan (x + 1) + \frac{7}{4} sin (2 arctan (x + 1)) - \frac{2}{x ^{2} + 2 x + 2} - \frac{2 ( x + 1 )}{x ^{2} + 2 x + 2} + C

解答ステップ

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

以下の部分分数を得る:

\frac{x ^{3} - x ^{2} + 1}{( x ^{2} + 2 x + 2 ) ^{2}} : \frac{x - 3}{x ^{2} + 2 x + 2} + \frac{4 x + 7}{( x ^{2} + 2 x + 2 ) ^{2}}

= \int \frac{x - 3}{x ^{2} + 2 x + 2} + \frac{4 x + 7}{( x ^{2} + 2 x + 2 ) ^{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{x - 3}{x ^{2} + 2 x + 2} d x + \int \frac{4 x + 7}{( x ^{2} + 2 x + 2 ) ^{2}} d x

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

\int \frac{4 x + 7}{( x ^{2} + 2 x + 2 ) ^{2}} d x = \frac{3}{2} arctan (x + 1) + \frac{7}{4} sin (2 arctan (x + 1)) - \frac{2}{x ^{2} + 2 x + 2} - \frac{2 ( x + 1 )}{x ^{2} + 2 x + 2}

= \frac{1}{2} ln x^{2} + 2 x + 2 - 4 arctan (x + 1) + \frac{3}{2} arctan (x + 1) + \frac{7}{4} sin (2 arctan (x + 1)) - \frac{2}{x ^{2} + 2 x + 2} - \frac{2 ( x + 1 )}{x ^{2} + 2 x + 2}

類似した元を足す:

- 4 arctan (x + 1) + \frac{3}{2} arctan (x + 1) = - \frac{5}{2} arctan (x + 1)

= \frac{1}{2} ln x^{2} + 2 x + 2 - \frac{5}{2} arctan (x + 1) + \frac{7}{4} sin (2 arctan (x + 1)) - \frac{2}{x ^{2} + 2 x + 2} - \frac{2 ( x + 1 )}{x ^{2} + 2 x + 2}

定数を解答に追加する

= \frac{1}{2} ln x^{2} + 2 x + 2 - \frac{5}{2} arctan (x + 1) + \frac{7}{4} sin (2 arctan (x + 1)) - \frac{2}{x ^{2} + 2 x + 2} - \frac{2 ( x + 1 )}{x ^{2} + 2 x + 2} + C