integral of (3x+1)/(x^3+4x^2+5x)

解

\int \frac{3 x + 1}{x ^{3} + 4 x ^{2} + 5 x} d x

\frac{1}{5} ln ∣ x ∣ + \frac{13}{5} arctan (x + 2) - \frac{1}{10} ln x^{2} + 4 x + 5 + C

解答ステップ

\int \frac{3 x + 1}{x ^{3} + 4 x ^{2} + 5 x} d x

拡張

\frac{3 x + 1}{x ^{3} + 4 x ^{2} + 5 x} : \frac{3 x}{x ^{3} + 4 x ^{2} + 5 x} + \frac{1}{x ^{3} + 4 x ^{2} + 5 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{3 x}{x ^{3} + 4 x ^{2} + 5 x} d x + \int \frac{1}{x ^{3} + 4 x ^{2} + 5 x} d x

\int \frac{3 x}{x ^{3} + 4 x ^{2} + 5 x} d x = 3 arctan (x + 2)

\int \frac{1}{x ^{3} + 4 x ^{2} + 5 x} d x = \frac{1}{5} ln ∣ x ∣ + \frac{1}{10} (- ln x^{2} + 4 x + 5 - 4 arctan (x + 2))

= 3 arctan (x + 2) + \frac{1}{5} ln ∣ x ∣ + \frac{1}{10} (- ln x^{2} + 4 x + 5 - 4 arctan (x + 2))

簡素化

3 arctan (x + 2) + \frac{1}{5} ln ∣ x ∣ + \frac{1}{10} (- ln x^{2} + 4 x + 5 - 4 arctan (x + 2)) : \frac{1}{5} ln ∣ x ∣ + \frac{13}{5} arctan (x + 2) - \frac{1}{10} ln x^{2} + 4 x + 5

= \frac{1}{5} ln ∣ x ∣ + \frac{13}{5} arctan (x + 2) - \frac{1}{10} ln x^{2} + 4 x + 5

定数を解答に追加する

= \frac{1}{5} ln ∣ x ∣ + \frac{13}{5} arctan (x + 2) - \frac{1}{10} ln x^{2} + 4 x + 5 + C