integral of (x^3)/(1-9x^2)

解

\int \frac{x ^{3}}{1 - 9 x ^{2}} d x

- \frac{x ^{2}}{18} - \frac{1}{162} ln - 9 x^{2} + 1 + C

解答ステップ

\int \frac{x ^{3}}{1 - 9 x ^{2}} d x

長除法

\frac{x ^{3}}{1 - 9 x ^{2}} : - \frac{x}{9} + \frac{x}{9 ( - 9 x ^{2} + 1 )}

= \int - \frac{x}{9} + \frac{x}{9 ( - 9 x ^{2} + 1 )} 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{x}{9} d x + \int \frac{x}{9 ( - 9 x ^{2} + 1 )} d x

\int \frac{x}{9} d x = \frac{x ^{2}}{18}

\int \frac{x}{9 ( - 9 x ^{2} + 1 )} d x = - \frac{1}{162} ln - 9 x^{2} + 1

= - \frac{x ^{2}}{18} - \frac{1}{162} ln - 9 x^{2} + 1

定数を解答に追加する

= - \frac{x ^{2}}{18} - \frac{1}{162} ln - 9 x^{2} + 1 + C