integral of 1+1/x

解答

\int 1 + \frac{1}{x} d x

x + ln ∣ x ∣ + C

求解步骤

\int 1 + \frac{1}{x} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d x + \int \frac{1}{x} d x

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

= x + ln ∣ x ∣

解答补常数

= x + ln ∣ x ∣ + C

\frac{\partial}{\partial y} (5 x^{2} + 8 x y + y^{2})

\frac{d y}{d x} + \frac{5 x ^{2}}{e ^{3 x^{3}}} y = 0

x y^{^{'}} - 2 y = 2 x^{2} + x - 2

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

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