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

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

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

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

定数を解答に追加する

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