integral of (x^3+1)/((x^2-4x+5)^2)

解

\int \frac{x ^{3} + 1}{( x ^{2} - 4 x + 5 ) ^{2}} d x

\frac{1}{2} ln x^{2} - 4 x + 5 + \frac{15}{2} arctan (x - 2) - \frac{19}{4} sin (2 arctan (x - 2)) - \frac{11}{2 ( x ^{2} - 4 x + 5 )} + \frac{11 ( x - 2 )}{x ^{2} - 4 x + 5} + C

解答ステップ

\int \frac{x ^{3} + 1}{( x ^{2} - 4 x + 5 ) ^{2}} d x

以下の部分分数を得る:

\frac{x ^{3} + 1}{( x ^{2} - 4 x + 5 ) ^{2}} : \frac{x + 4}{x ^{2} - 4 x + 5} + \frac{11 x - 19}{( x ^{2} - 4 x + 5 ) ^{2}}

= \int \frac{x + 4}{x ^{2} - 4 x + 5} + \frac{11 x - 19}{( x ^{2} - 4 x + 5 ) ^{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 + 4}{x ^{2} - 4 x + 5} d x + \int \frac{11 x - 19}{( x ^{2} - 4 x + 5 ) ^{2}} d x

\int \frac{x + 4}{x ^{2} - 4 x + 5} d x = \frac{1}{2} ln x^{2} - 4 x + 5 + 6 arctan (x - 2)

\int \frac{11 x - 19}{( x ^{2} - 4 x + 5 ) ^{2}} d x = \frac{3}{2} arctan (x - 2) - \frac{19}{4} sin (2 arctan (x - 2)) - \frac{11}{2 ( x ^{2} - 4 x + 5 )} + \frac{11 ( x - 2 )}{x ^{2} - 4 x + 5}

= \frac{1}{2} ln x^{2} - 4 x + 5 + 6 arctan (x - 2) + \frac{3}{2} arctan (x - 2) - \frac{19}{4} sin (2 arctan (x - 2)) - \frac{11}{2 ( x ^{2} - 4 x + 5 )} + \frac{11 ( x - 2 )}{x ^{2} - 4 x + 5}

類似した元を足す:

6 arctan (x - 2) + \frac{3}{2} arctan (x - 2) = \frac{15}{2} arctan (x - 2)

= \frac{1}{2} ln x^{2} - 4 x + 5 + \frac{15}{2} arctan (x - 2) - \frac{19}{4} sin (2 arctan (x - 2)) - \frac{11}{2 ( x ^{2} - 4 x + 5 )} + \frac{11 ( x - 2 )}{x ^{2} - 4 x + 5}

定数を解答に追加する

= \frac{1}{2} ln x^{2} - 4 x + 5 + \frac{15}{2} arctan (x - 2) - \frac{19}{4} sin (2 arctan (x - 2)) - \frac{11}{2 ( x ^{2} - 4 x + 5 )} + \frac{11 ( x - 2 )}{x ^{2} - 4 x + 5} + C