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

解答

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

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

求解步骤

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

乘开

x (1 + \frac{1}{x})^{3} : x + 3 + \frac{3}{x} + \frac{1}{x ^{2}}

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

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

\int 3 d x = 3 x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

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

解答补常数

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

x \to + 1 lim (ln (3) + e^{x})

\int \frac{2 x ^{2} + 24}{x ( x - 2 ) ^{3}} d x

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

\int \frac{1}{a ^{2} - u ^{2}} d u

θ^{^{''}} + \frac{a}{b} \cdot θ = 0