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

解

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

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

解答ステップ

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

拡張

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

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

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

\int \frac{2}{x ^{2} ( x ^{2} + 4 )} d x = 2 (- \frac{1}{4 x} - \frac{1}{8} arctan (\frac{x}{2}))

= \frac{1}{2} arctan (\frac{x}{2}) + 2 (- \frac{1}{4 x} - \frac{1}{8} arctan (\frac{x}{2}))

簡素化

\frac{1}{2} arctan (\frac{x}{2}) + 2 (- \frac{1}{4 x} - \frac{1}{8} arctan (\frac{x}{2})) : \frac{1}{4} arctan (\frac{x}{2}) - \frac{1}{2 x}

= \frac{1}{4} arctan (\frac{x}{2}) - \frac{1}{2 x}

定数を解答に追加する

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