integral of (6+t+t^2)/(sqrt(t))

解

\int \frac{6 + t + t ^{2}}{t} d t

12 t + \frac{2}{3} t^{\frac{3}{2}} + \frac{2}{5} t^{\frac{5}{2}} + C

解答ステップ

\int \frac{6 + t + t ^{2}}{t} d t

拡張

\frac{6 + t + t ^{2}}{t} : \frac{6}{t} + t + t^{\frac{3}{2}}

= \int \frac{6}{t} + t + t^{\frac{3}{2}} 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{6}{t} d t + \int t d t + \int t^{\frac{3}{2}} d t

\int \frac{6}{t} d t = 12 t

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

\int t^{\frac{3}{2}} d t = \frac{2}{5} t^{\frac{5}{2}}

= 12 t + \frac{2}{3} t^{\frac{3}{2}} + \frac{2}{5} t^{\frac{5}{2}}

定数を解答に追加する

= 12 t + \frac{2}{3} t^{\frac{3}{2}} + \frac{2}{5} t^{\frac{5}{2}} + C