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

解

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

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

解答ステップ

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

拡張

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

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

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

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

簡素化

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

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

定数を解答に追加する

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