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

解

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

4 ln ∣ x ∣ - \frac{1}{2 x ^{2}} + C

解答ステップ

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

拡張

\frac{4 x ^{2} + 1}{x ^{3}} : \frac{4}{x} + \frac{1}{x ^{3}}

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

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

= 4 ln ∣ x ∣ - \frac{1}{2 x ^{2}}

定数を解答に追加する

= 4 ln ∣ x ∣ - \frac{1}{2 x ^{2}} + C