integral of ((1-u^2-u-u^3))/(1+u^2)

解

\int \frac{( 1 - u ^{2} - u - u ^{3} )}{1 + u ^{2}} d u

- \frac{1}{2} u^{2} - u + 2 arctan (u) - \frac{1}{2} + C

解答ステップ

\int \frac{1 - u ^{2} - u - u ^{3}}{1 + u ^{2}} d u

拡張

\frac{1 - u ^{2} - u - u ^{3}}{1 + u ^{2}} : \frac{1}{1 + u ^{2}} - \frac{u ^{2}}{1 + u ^{2}} - \frac{u}{1 + u ^{2}} - \frac{u ^{3}}{1 + u ^{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{1}{1 + u ^{2}} d u - \int \frac{u ^{2}}{1 + u ^{2}} d u - \int \frac{u}{1 + u ^{2}} d u - \int \frac{u ^{3}}{1 + u ^{2}} d u

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

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

\int \frac{u}{1 + u ^{2}} d u = \frac{1}{2} ln 1 + u^{2}

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

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

簡素化

arctan (u) - (- arctan (u) + u) - \frac{1}{2} ln 1 + u^{2} - \frac{1}{2} (1 + u^{2} - ln 1 + u^{2}) : - \frac{1}{2} u^{2} - u + 2 arctan (u) - \frac{1}{2}

= - \frac{1}{2} u^{2} - u + 2 arctan (u) - \frac{1}{2}

定数を解答に追加する

= - \frac{1}{2} u^{2} - u + 2 arctan (u) - \frac{1}{2} + C