integral of (x+1)/(x^4+3x^2)

解

\int \frac{x + 1}{x ^{4} + 3 x ^{2}} d x

\frac{1}{3} ln (x) - \frac{1}{6} ln x^{2} + 3 - \frac{1}{3 x} - \frac{1}{3 3} arctan (\frac{x}{3}) + C

解答ステップ

\int \frac{x + 1}{x ^{4} + 3 x ^{2}} d x

拡張

\frac{x + 1}{x ^{4} + 3 x ^{2}} : \frac{x}{x ^{4} + 3 x ^{2}} + \frac{1}{x ^{4} + 3 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{x}{x ^{4} + 3 x ^{2}} d x + \int \frac{1}{x ^{4} + 3 x ^{2}} d x

\int \frac{x}{x ^{4} + 3 x ^{2}} d x = \frac{1}{3} ln (x) - \frac{1}{6} ln x^{2} + 3

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

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

定数を解答に追加する

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