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

解

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

3 t^{\frac{5}{3}} - \frac{21}{t ^{\frac{1}{3}}} + C

解答ステップ

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

拡張

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

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

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

\int \frac{7}{t ^{\frac{4}{3}}} d t = - \frac{21}{t ^{\frac{1}{3}}}

= 3 t^{\frac{5}{3}} - \frac{21}{t ^{\frac{1}{3}}}

定数を解答に追加する

= 3 t^{\frac{5}{3}} - \frac{21}{t ^{\frac{1}{3}}} + C