integral of (5x^3+7x+2)/(x^4+9x^2)

解

\int \frac{5 x ^{3} + 7 x + 2}{x ^{4} + 9 x ^{2}} d x

\frac{19}{9} ln x^{2} + 9 + \frac{7}{9} ln (x) - \frac{2}{27} arctan (\frac{x}{3}) - \frac{2}{9 x} + C

解答ステップ

\int \frac{5 x ^{3} + 7 x + 2}{x ^{4} + 9 x ^{2}} d x

拡張

\frac{5 x ^{3} + 7 x + 2}{x ^{4} + 9 x ^{2}} : \frac{5 x ^{3}}{x ^{4} + 9 x ^{2}} + \frac{7 x}{x ^{4} + 9 x ^{2}} + \frac{2}{x ^{4} + 9 x ^{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{5 x ^{3}}{x ^{4} + 9 x ^{2}} d x + \int \frac{7 x}{x ^{4} + 9 x ^{2}} d x + \int \frac{2}{x ^{4} + 9 x ^{2}} d x

\int \frac{5 x ^{3}}{x ^{4} + 9 x ^{2}} d x = \frac{5}{2} ln x^{2} + 9

\int \frac{7 x}{x ^{4} + 9 x ^{2}} d x = 7 (\frac{1}{9} ln (x) - \frac{1}{18} ln x^{2} + 9)

\int \frac{2}{x ^{4} + 9 x ^{2}} d x = 2 (- \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3}))

= \frac{5}{2} ln x^{2} + 9 + 7 (\frac{1}{9} ln (x) - \frac{1}{18} ln x^{2} + 9) + 2 (- \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3}))

簡素化

\frac{5}{2} ln x^{2} + 9 + 7 (\frac{1}{9} ln (x) - \frac{1}{18} ln x^{2} + 9) + 2 (- \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3})) : \frac{19}{9} ln x^{2} + 9 + \frac{7}{9} ln (x) - \frac{2}{27} arctan (\frac{x}{3}) - \frac{2}{9 x}

= \frac{19}{9} ln x^{2} + 9 + \frac{7}{9} ln (x) - \frac{2}{27} arctan (\frac{x}{3}) - \frac{2}{9 x}

定数を解答に追加する

= \frac{19}{9} ln x^{2} + 9 + \frac{7}{9} ln (x) - \frac{2}{27} arctan (\frac{x}{3}) - \frac{2}{9 x} + C