integral of (4x^3+2x^2+8)/((x^2+4)^2)

解

\int \frac{4 x ^{3} + 2 x ^{2} + 8}{( x ^{2} + 4 ) ^{2}} d x

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

解答ステップ

\int \frac{4 x ^{3} + 2 x ^{2} + 8}{( x ^{2} + 4 ) ^{2}} d x

拡張

\frac{4 x ^{3} + 2 x ^{2} + 8}{( x ^{2} + 4 ) ^{2}} : \frac{4 x ^{3}}{( x ^{2} + 4 ) ^{2}} + \frac{2 x ^{2}}{( x ^{2} + 4 ) ^{2}} + \frac{8}{( 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 x ^{3}}{( x ^{2} + 4 ) ^{2}} d x + \int \frac{2 x ^{2}}{( x ^{2} + 4 ) ^{2}} d x + \int \frac{8}{( x ^{2} + 4 ) ^{2}} d x

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

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

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

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

簡素化

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

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

定数を解答に追加する

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