integral of (5t^2+7)/(t^{3/4)}

解

\int \frac{5 t ^{2} + 7}{t ^{\frac{3}{4}}} d t

\frac{20}{9} t^{\frac{9}{4}} + 28 t^{\frac{1}{4}} + C

解答ステップ

\int \frac{5 t ^{2} + 7}{t ^{\frac{3}{4}}} d t

拡張

\frac{5 t ^{2} + 7}{t ^{\frac{3}{4}}} : 5 t^{\frac{5}{4}} + \frac{7}{t ^{\frac{3}{4}}}

= \int 5 t^{\frac{5}{4}} + \frac{7}{t ^{\frac{3}{4}}} 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 5 t^{\frac{5}{4}} d t + \int \frac{7}{t ^{\frac{3}{4}}} d t

\int 5 t^{\frac{5}{4}} d t = \frac{20}{9} t^{\frac{9}{4}}

\int \frac{7}{t ^{\frac{3}{4}}} d t = 28 t^{\frac{1}{4}}

= \frac{20}{9} t^{\frac{9}{4}} + 28 t^{\frac{1}{4}}

定数を解答に追加する

= \frac{20}{9} t^{\frac{9}{4}} + 28 t^{\frac{1}{4}} + C