integral of ((t^3-8t+1)/((2t)^4))

解

\int (\frac{t ^{3} - 8 t + 1}{( 2 t ) ^{4}}) d t

\frac{1}{16} ln ∣ t ∣ + \frac{1}{4 t ^{2}} - \frac{1}{48 t ^{3}} + C

解答ステップ

\int \frac{t ^{3} - 8 t + 1}{( 2 t ) ^{4}} d t

拡張

\frac{t ^{3} - 8 t + 1}{( 2 t ) ^{4}} : \frac{1}{16 t} - \frac{1}{2 t ^{3}} + \frac{1}{16 t ^{4}}

= \int \frac{1}{16 t} - \frac{1}{2 t ^{3}} + \frac{1}{16 t ^{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 \frac{1}{16 t} d t - \int \frac{1}{2 t ^{3}} d t + \int \frac{1}{16 t ^{4}} d t

\int \frac{1}{16 t} d t = \frac{1}{16} ln ∣ t ∣

\int \frac{1}{2 t ^{3}} d t = - \frac{1}{4 t ^{2}}

\int \frac{1}{16 t ^{4}} d t = - \frac{1}{48 t ^{3}}

= \frac{1}{16} ln ∣ t ∣ - (- \frac{1}{4 t ^{2}}) - \frac{1}{48 t ^{3}}

簡素化

= \frac{1}{16} ln ∣ t ∣ + \frac{1}{4 t ^{2}} - \frac{1}{48 t ^{3}}

定数を解答に追加する

= \frac{1}{16} ln ∣ t ∣ + \frac{1}{4 t ^{2}} - \frac{1}{48 t ^{3}} + C