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

解

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

\frac{9}{68} ln x^{2} + 9 + \frac{29}{34} ln ∣ x - 5 ∣ + \frac{15}{34} arctan (\frac{x}{3}) + \frac{1}{51} (- 3 ln x^{2} + 9 - 10 arctan (\frac{x}{3})) + C

解答ステップ

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

拡張

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

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

\int \frac{x ^{2}}{( x - 5 ) ( x ^{2} + 9 )} d x = \frac{25}{34} ln ∣ x - 5 ∣ + \frac{1}{34} (\frac{9}{2} ln x^{2} + 9 + 15 arctan (\frac{x}{3}))

\int \frac{4}{( x - 5 ) ( x ^{2} + 9 )} d x = 4 (\frac{1}{34} ln ∣ x - 5 ∣ + \frac{1}{34} (- \frac{1}{2} ln x^{2} + 9 - \frac{5}{3} arctan (\frac{x}{3})))

= \frac{25}{34} ln ∣ x - 5 ∣ + \frac{1}{34} (\frac{9}{2} ln x^{2} + 9 + 15 arctan (\frac{x}{3})) + 4 (\frac{1}{34} ln ∣ x - 5 ∣ + \frac{1}{34} (- \frac{1}{2} ln x^{2} + 9 - \frac{5}{3} arctan (\frac{x}{3})))

簡素化

\frac{25}{34} ln ∣ x - 5 ∣ + \frac{1}{34} (\frac{9}{2} ln x^{2} + 9 + 15 arctan (\frac{x}{3})) + 4 (\frac{1}{34} ln ∣ x - 5 ∣ + \frac{1}{34} (- \frac{1}{2} ln x^{2} + 9 - \frac{5}{3} arctan (\frac{x}{3}))) : \frac{9}{68} ln x^{2} + 9 + \frac{29}{34} ln ∣ x - 5 ∣ + \frac{15}{34} arctan (\frac{x}{3}) + \frac{1}{51} (- 3 ln x^{2} + 9 - 10 arctan (\frac{x}{3}))

= \frac{9}{68} ln x^{2} + 9 + \frac{29}{34} ln ∣ x - 5 ∣ + \frac{15}{34} arctan (\frac{x}{3}) + \frac{1}{51} (- 3 ln x^{2} + 9 - 10 arctan (\frac{x}{3}))

定数を解答に追加する

= \frac{9}{68} ln x^{2} + 9 + \frac{29}{34} ln ∣ x - 5 ∣ + \frac{15}{34} arctan (\frac{x}{3}) + \frac{1}{51} (- 3 ln x^{2} + 9 - 10 arctan (\frac{x}{3})) + C