integral of (10)/((3x-3)(x^2+9))

解

\int \frac{10}{( 3 x - 3 ) ( x ^{2} + 9 )} d x

10 (\frac{1}{30} ln ∣ x - 1 ∣ + \frac{1}{30} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}))) + C

解答ステップ

\int \frac{10}{( 3 x - 3 ) ( x ^{2} + 9 )} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 \cdot \int \frac{1}{( 3 x - 3 ) ( x ^{2} + 9 )} d x

以下の部分分数を得る:

\frac{1}{( 3 x - 3 ) ( x ^{2} + 9 )} : \frac{1}{30 ( x - 1 )} + \frac{- x - 1}{30 ( x ^{2} + 9 )}

= 10 \cdot \int \frac{1}{30 ( x - 1 )} + \frac{- x - 1}{30 ( 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

= 10 (\int \frac{1}{30 ( x - 1 )} d x + \int \frac{- x - 1}{30 ( x ^{2} + 9 )} d x)

\int \frac{1}{30 ( x - 1 )} d x = \frac{1}{30} ln ∣ x - 1 ∣

\int \frac{- x - 1}{30 ( x ^{2} + 9 )} d x = \frac{1}{30} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}))

= 10 (\frac{1}{30} ln ∣ x - 1 ∣ + \frac{1}{30} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3})))

定数を解答に追加する

= 10 (\frac{1}{30} ln ∣ x - 1 ∣ + \frac{1}{30} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}))) + C