integral of x^3-4/(\sqrt[3]{x)}+9/x+6e^x

解答

\int x^{3} - \frac{4}{3 x} + \frac{9}{x} + 6 e^{x} d x

\frac{x ^{4}}{4} - 6 x^{\frac{2}{3}} + 9 ln ∣ x ∣ + 6 e^{x} + C

求解步骤

\int x^{3} - \frac{4}{3 x} + \frac{9}{x} + 6 e^{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 x^{3} d x - \int \frac{4}{3 x} d x + \int \frac{9}{x} d x + \int 6 e^{x} d x

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

\int \frac{4}{3 x} d x = 6 x^{\frac{2}{3}}

\int \frac{9}{x} d x = 9 ln ∣ x ∣

\int 6 e^{x} d x = 6 e^{x}

= \frac{x ^{4}}{4} - 6 x^{\frac{2}{3}} + 9 ln ∣ x ∣ + 6 e^{x}

解答补常数

= \frac{x ^{4}}{4} - 6 x^{\frac{2}{3}} + 9 ln ∣ x ∣ + 6 e^{x} + C

in v erse l a pl a ce \frac{1}{( s ^{2} - 9 )}

\frac{d y}{d x} + 2 x^{2} y = 11 x^{2}

\int \frac{e ^{2 x} + 1}{e ^{2 x} - 1} d x

x \to 0 - lim (1 + \frac{1}{x})

\int_{- \infty}^{6} 10 x e^{2 x} d x