integral of (3/(u^2)+4/(u^3)-2)

解

\int (\frac{3}{u ^{2}} + \frac{4}{u ^{3}} - 2) d u

- \frac{3}{u} - \frac{2}{u ^{2}} - 2 u + C

解答ステップ

\int \frac{3}{u ^{2}} + \frac{4}{u ^{3}} - 2 d u

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{3}{u ^{2}} d u + \int \frac{4}{u ^{3}} d u - \int 2 d u

\int \frac{3}{u ^{2}} d u = - \frac{3}{u}

\int \frac{4}{u ^{3}} d u = - \frac{2}{u ^{2}}

\int 2 d u = 2 u

= - \frac{3}{u} - \frac{2}{u ^{2}} - 2 u

定数を解答に追加する

= - \frac{3}{u} - \frac{2}{u ^{2}} - 2 u + C