integral of (4x+6)/((x^2+2)*(x+2))

解

\int \frac{4 x + 6}{( x ^{2} + 2 ) \cdot ( x + 2 )} d x

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

解答ステップ

\int \frac{4 x + 6}{( x ^{2} + 2 ) ( x + 2 )} d x

以下の部分分数を得る:

\frac{4 x + 6}{( x ^{2} + 2 ) ( x + 2 )} : \frac{x + 10}{3 ( x ^{2} + 2 )} - \frac{1}{3 ( x + 2 )}

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

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

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

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

定数を解答に追加する

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