integral of (4x^2+29)/(x^3+6x^2+9x-50)

解

\int \frac{4 x ^{2} + 29}{x ^{3} + 6 x ^{2} + 9 x - 50} d x

ln ∣ x - 2 ∣ - \frac{29}{90} ln x^{2} + 8 x + 25 - \frac{14}{3} arctan (\frac{1}{3} (x + 4)) + \frac{82}{45} ln \frac{1}{9} (x + 4)^{2} + 1 + C

解答ステップ

\int \frac{4 x ^{2} + 29}{x ^{3} + 6 x ^{2} + 9 x - 50} d x

拡張

\frac{4 x ^{2} + 29}{x ^{3} + 6 x ^{2} + 9 x - 50} : \frac{4 x ^{2}}{x ^{3} + 6 x ^{2} + 9 x - 50} + \frac{29}{x ^{3} + 6 x ^{2} + 9 x - 50}

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 ^{2}}{x ^{3} + 6 x ^{2} + 9 x - 50} d x + \int \frac{29}{x ^{3} + 6 x ^{2} + 9 x - 50} d x

\int \frac{4 x ^{2}}{x ^{3} + 6 x ^{2} + 9 x - 50} d x = 4 (\frac{4}{45} ln ∣ x - 2 ∣ + \frac{1}{45} (\frac{41}{2} ln \frac{1}{9} (x + 4)^{2} + 1 - 38 arctan (\frac{1}{3} (x + 4))))

\int \frac{29}{x ^{3} + 6 x ^{2} + 9 x - 50} d x = 29 (\frac{1}{45} ln ∣ x - 2 ∣ - \frac{1}{90} (ln x^{2} + 8 x + 25 + 4 arctan (\frac{1}{3} (x + 4))))

= 4 (\frac{4}{45} ln ∣ x - 2 ∣ + \frac{1}{45} (\frac{41}{2} ln \frac{1}{9} (x + 4)^{2} + 1 - 38 arctan (\frac{1}{3} (x + 4)))) + 29 (\frac{1}{45} ln ∣ x - 2 ∣ - \frac{1}{90} (ln x^{2} + 8 x + 25 + 4 arctan (\frac{1}{3} (x + 4))))

簡素化

4 (\frac{4}{45} ln ∣ x - 2 ∣ + \frac{1}{45} (\frac{41}{2} ln \frac{1}{9} (x + 4)^{2} + 1 - 38 arctan (\frac{1}{3} (x + 4)))) + 29 (\frac{1}{45} ln ∣ x - 2 ∣ - \frac{1}{90} (ln x^{2} + 8 x + 25 + 4 arctan (\frac{1}{3} (x + 4)))) : ln ∣ x - 2 ∣ - \frac{29}{90} ln x^{2} + 8 x + 25 - \frac{14}{3} arctan (\frac{1}{3} (x + 4)) + \frac{82}{45} ln \frac{1}{9} (x + 4)^{2} + 1

= ln ∣ x - 2 ∣ - \frac{29}{90} ln x^{2} + 8 x + 25 - \frac{14}{3} arctan (\frac{1}{3} (x + 4)) + \frac{82}{45} ln \frac{1}{9} (x + 4)^{2} + 1

定数を解答に追加する

= ln ∣ x - 2 ∣ - \frac{29}{90} ln x^{2} + 8 x + 25 - \frac{14}{3} arctan (\frac{1}{3} (x + 4)) + \frac{82}{45} ln \frac{1}{9} (x + 4)^{2} + 1 + C