integral of (bx^5+1)/(x^3+6x^2+9x)

解

\int \frac{b x ^{5} + 1}{x ^{3} + 6 x ^{2} + 9 x} d x

b (\frac{x ^{3}}{3} - 3 x^{2} + 27 x - 27 (4 ln ∣ x + 3 ∣ + \frac{3}{x + 3})) + \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x + 3 ∣ + \frac{1}{3 ( x + 3 )} + C

解答ステップ

\int \frac{b x ^{5} + 1}{x ^{3} + 6 x ^{2} + 9 x} d x

拡張

\frac{b x ^{5} + 1}{x ^{3} + 6 x ^{2} + 9 x} : \frac{b x ^{5}}{x ^{3} + 6 x ^{2} + 9 x} + \frac{1}{x ^{3} + 6 x ^{2} + 9 x}

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{b x ^{5}}{x ^{3} + 6 x ^{2} + 9 x} d x + \int \frac{1}{x ^{3} + 6 x ^{2} + 9 x} d x

\int \frac{b x ^{5}}{x ^{3} + 6 x ^{2} + 9 x} d x = b (\frac{x ^{3}}{3} - 3 x^{2} + 27 x - 27 (4 ln ∣ x + 3 ∣ + \frac{3}{x + 3}))

\int \frac{1}{x ^{3} + 6 x ^{2} + 9 x} d x = \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x + 3 ∣ + \frac{1}{3 ( x + 3 )}

= b (\frac{x ^{3}}{3} - 3 x^{2} + 27 x - 27 (4 ln ∣ x + 3 ∣ + \frac{3}{x + 3})) + \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x + 3 ∣ + \frac{1}{3 ( x + 3 )}

定数を解答に追加する

= b (\frac{x ^{3}}{3} - 3 x^{2} + 27 x - 27 (4 ln ∣ x + 3 ∣ + \frac{3}{x + 3})) + \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x + 3 ∣ + \frac{1}{3 ( x + 3 )} + C