integral of (6x^2+1)/(8x)

解

\int \frac{6 x ^{2} + 1}{8 x} d x

\frac{1}{8} (3 x^{2} + ln ∣ x ∣) + C

解答ステップ

\int \frac{6 x ^{2} + 1}{8 x} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int \frac{6 x ^{2} + 1}{x} d x

拡張

\frac{6 x ^{2} + 1}{x} : 6 x + \frac{1}{x}

= \frac{1}{8} \cdot \int 6 x + \frac{1}{x} 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

= \frac{1}{8} (\int 6 x d x + \int \frac{1}{x} d x)

\int 6 x d x = 3 x^{2}

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{1}{8} (3 x^{2} + ln ∣ x ∣)

定数を解答に追加する

= \frac{1}{8} (3 x^{2} + ln ∣ x ∣) + C