integral of (1+1/t)(1/(t^2))

解

\int (1 + \frac{1}{t}) (\frac{1}{t ^{2}}) d t

- \frac{1}{t} - \frac{1}{2 t ^{2}} + C

解答ステップ

\int (1 + \frac{1}{t}) \frac{1}{t ^{2}} d t

簡素化

(1 + \frac{1}{t}) \frac{1}{t ^{2}} : \frac{t + 1}{t ^{3}}

= \int \frac{t + 1}{t ^{3}} d t

拡張

\frac{t + 1}{t ^{3}} : \frac{1}{t ^{2}} + \frac{1}{t ^{3}}

= \int \frac{1}{t ^{2}} + \frac{1}{t ^{3}} d t

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}{t ^{2}} d t + \int \frac{1}{t ^{3}} d t

\int \frac{1}{t ^{2}} d t = - \frac{1}{t}

\int \frac{1}{t ^{3}} d t = - \frac{1}{2 t ^{2}}

= - \frac{1}{t} - \frac{1}{2 t ^{2}}

定数を解答に追加する

= - \frac{1}{t} - \frac{1}{2 t ^{2}} + C