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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

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

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

簡素化

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

定数を解答に追加する

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