integral of (-x^2-4x+18)/((3x+4)(x^2+9))

解

\int \frac{- x ^{2} - 4 x + 18}{( 3 x + 4 ) ( x ^{2} + 9 )} d x

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

解答ステップ

\int \frac{- x ^{2} - 4 x + 18}{( 3 x + 4 ) ( x ^{2} + 9 )} d x

以下の部分分数を得る:

\frac{- x ^{2} - 4 x + 18}{( 3 x + 4 ) ( x ^{2} + 9 )} : \frac{2}{3 x + 4} - \frac{x}{x ^{2} + 9}

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

\int \frac{2}{3 x + 4} d x = \frac{2}{3} ln ∣ 3 x + 4 ∣

\int \frac{x}{x ^{2} + 9} d x = \frac{1}{2} ln x^{2} + 9

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

定数を解答に追加する

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