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

解

\int \frac{4 x - 2}{x ^{4} + x ^{2}} d x

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

解答ステップ

\int \frac{4 x - 2}{x ^{4} + x ^{2}} d x

拡張

\frac{4 x - 2}{x ^{4} + x ^{2}} : \frac{4 x}{x ^{4} + x ^{2}} - \frac{2}{x ^{4} + x ^{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{4 x}{x ^{4} + x ^{2}} d x - \int \frac{2}{x ^{4} + x ^{2}} d x

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

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

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

定数を解答に追加する

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