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

解

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

\frac{3 x ^{2}}{2} - \frac{x ^{3}}{6} + sin (x) + ln ∣ x ∣ + C

解答ステップ

\int 3 x - \frac{1}{2} x^{2} + cos (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

= \int 3 x d x - \int \frac{1}{2} x^{2} d x + \int cos (x) d x + \int \frac{1}{x} d x

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

\int \frac{1}{2} x^{2} d x = \frac{x ^{3}}{6}

\int cos (x) d x = sin (x)

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{3 x ^{2}}{2} - \frac{x ^{3}}{6} + sin (x) + ln ∣ x ∣

定数を解答に追加する

= \frac{3 x ^{2}}{2} - \frac{x ^{3}}{6} + sin (x) + ln ∣ x ∣ + C