integral of 1t^3sqrt(t)+1/(tsqrt(t))

解

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

\frac{2}{9} t^{\frac{9}{2}} - \frac{2}{t} + C

解答ステップ

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

\int 1 \cdot t^{3} t d t = \frac{2}{9} t^{\frac{9}{2}}

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

= \frac{2}{9} t^{\frac{9}{2}} - \frac{2}{t}

定数を解答に追加する

= \frac{2}{9} t^{\frac{9}{2}} - \frac{2}{t} + C