integral of (8x^2+5x+18)/((4x^2+9)^2)

解

\int \frac{8 x ^{2} + 5 x + 18}{( 4 x ^{2} + 9 ) ^{2}} d x

\frac{1}{3} arctan (\frac{2}{3} x) + \frac{1}{12} sin (2 arctan (\frac{2}{3} x)) - \frac{x}{4 x ^{2} + 9} - \frac{5}{8 ( 4 x ^{2} + 9 )} + C

解答ステップ

\int \frac{8 x ^{2} + 5 x + 18}{( 4 x ^{2} + 9 ) ^{2}} d x

拡張

\frac{8 x ^{2} + 5 x + 18}{( 4 x ^{2} + 9 ) ^{2}} : \frac{8 x ^{2}}{( 4 x ^{2} + 9 ) ^{2}} + \frac{5 x}{( 4 x ^{2} + 9 ) ^{2}} + \frac{18}{( 4 x ^{2} + 9 ) ^{2}}

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{8 x ^{2}}{( 4 x ^{2} + 9 ) ^{2}} d x + \int \frac{5 x}{( 4 x ^{2} + 9 ) ^{2}} d x + \int \frac{18}{( 4 x ^{2} + 9 ) ^{2}} d x

\int \frac{8 x ^{2}}{( 4 x ^{2} + 9 ) ^{2}} d x = \frac{1}{6} arctan (\frac{2}{3} x) - \frac{x}{9 + 4 x ^{2}}

\int \frac{5 x}{( 4 x ^{2} + 9 ) ^{2}} d x = - \frac{5}{8 ( 4 x ^{2} + 9 )}

\int \frac{18}{( 4 x ^{2} + 9 ) ^{2}} d x = \frac{1}{6} (arctan (\frac{2}{3} x) + \frac{1}{2} sin (2 arctan (\frac{2}{3} x)))

= \frac{1}{6} arctan (\frac{2}{3} x) - \frac{x}{9 + 4 x ^{2}} - \frac{5}{8 ( 4 x ^{2} + 9 )} + \frac{1}{6} (arctan (\frac{2}{3} x) + \frac{1}{2} sin (2 arctan (\frac{2}{3} x)))

簡素化

\frac{1}{6} arctan (\frac{2}{3} x) - \frac{x}{9 + 4 x ^{2}} - \frac{5}{8 ( 4 x ^{2} + 9 )} + \frac{1}{6} (arctan (\frac{2}{3} x) + \frac{1}{2} sin (2 arctan (\frac{2}{3} x))) : \frac{1}{3} arctan (\frac{2}{3} x) + \frac{1}{12} sin (2 arctan (\frac{2}{3} x)) - \frac{x}{4 x ^{2} + 9} - \frac{5}{8 ( 4 x ^{2} + 9 )}

= \frac{1}{3} arctan (\frac{2}{3} x) + \frac{1}{12} sin (2 arctan (\frac{2}{3} x)) - \frac{x}{4 x ^{2} + 9} - \frac{5}{8 ( 4 x ^{2} + 9 )}

定数を解答に追加する

= \frac{1}{3} arctan (\frac{2}{3} x) + \frac{1}{12} sin (2 arctan (\frac{2}{3} x)) - \frac{x}{4 x ^{2} + 9} - \frac{5}{8 ( 4 x ^{2} + 9 )} + C