integral of 1/((x+3)(9x^2+4))

解

\int \frac{1}{( x + 3 ) ( 9 x ^{2} + 4 )} d x

\frac{1}{85} ln ∣ x + 3 ∣ + \frac{1}{85} (- \frac{1}{2} ln 9 x^{2} + 4 + \frac{9}{2} arctan (\frac{3}{2} x)) + C

解答ステップ

\int \frac{1}{( x + 3 ) ( 9 x ^{2} + 4 )} d x

以下の部分分数を得る:

\frac{1}{( x + 3 ) ( 9 x ^{2} + 4 )} : \frac{1}{85 ( x + 3 )} + \frac{- 9 x + 27}{85 ( 9 x ^{2} + 4 )}

= \int \frac{1}{85 ( x + 3 )} + \frac{- 9 x + 27}{85 ( 9 x ^{2} + 4 )} 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{1}{85 ( x + 3 )} d x + \int \frac{- 9 x + 27}{85 ( 9 x ^{2} + 4 )} d x

\int \frac{1}{85 ( x + 3 )} d x = \frac{1}{85} ln ∣ x + 3 ∣

\int \frac{- 9 x + 27}{85 ( 9 x ^{2} + 4 )} d x = \frac{1}{85} (- \frac{1}{2} ln 9 x^{2} + 4 + \frac{9}{2} arctan (\frac{3}{2} x))

= \frac{1}{85} ln ∣ x + 3 ∣ + \frac{1}{85} (- \frac{1}{2} ln 9 x^{2} + 4 + \frac{9}{2} arctan (\frac{3}{2} x))

定数を解答に追加する

= \frac{1}{85} ln ∣ x + 3 ∣ + \frac{1}{85} (- \frac{1}{2} ln 9 x^{2} + 4 + \frac{9}{2} arctan (\frac{3}{2} x)) + C