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

解

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

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

解答ステップ

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

拡張

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

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

\int \frac{6 x}{x ^{2} ( x ^{2} + 2 )} d x = - 3 (\frac{1}{2} ln x^{2} + 2 - ln (x))

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

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

定数を解答に追加する

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