integral of (x^4)/(25+4x^2)

解

\int \frac{x ^{4}}{25 + 4 x ^{2}} d x

\frac{x ^{3}}{12} - \frac{25}{16} x + \frac{125}{32} arctan (\frac{2}{5} x) + C

解答ステップ

\int \frac{x ^{4}}{25 + 4 x ^{2}} d x

長除法

\frac{x ^{4}}{25 + 4 x ^{2}} : \frac{x ^{2}}{4} - \frac{25}{16} + \frac{625}{16 ( 4 x ^{2} + 25 )}

= \int \frac{x ^{2}}{4} - \frac{25}{16} + \frac{625}{16 ( 4 x ^{2} + 25 )} 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 ^{2}}{4} d x - \int \frac{25}{16} d x + \int \frac{625}{16 ( 4 x ^{2} + 25 )} d x

\int \frac{x ^{2}}{4} d x = \frac{x ^{3}}{12}

\int \frac{25}{16} d x = \frac{25}{16} x

\int \frac{625}{16 ( 4 x ^{2} + 25 )} d x = \frac{125}{32} arctan (\frac{2}{5} x)

= \frac{x ^{3}}{12} - \frac{25}{16} x + \frac{125}{32} arctan (\frac{2}{5} x)

定数を解答に追加する

= \frac{x ^{3}}{12} - \frac{25}{16} x + \frac{125}{32} arctan (\frac{2}{5} x) + C