integral of (1/4)^x

解答

\int (\frac{1}{4})^{x} d x

- \frac{1}{ln ( 2 ) \cdot 2 ^{2 x + 1}} + C

求解步骤

\int (\frac{1}{4})^{x} d x

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{( \frac{1}{4} ) ^{x}}{ln ( \frac{1}{4} )}

化简

\frac{( \frac{1}{4} ) ^{x}}{ln ( \frac{1}{4} )} : - \frac{1}{ln ( 2 ) \cdot 2 ^{2 x + 1}}

= - \frac{1}{ln ( 2 ) \cdot 2 ^{2 x + 1}}

解答补常数

= - \frac{1}{ln ( 2 ) \cdot 2 ^{2 x + 1}} + C

x \to - 3 lim (\frac{1 - 6 x}{3 + x})

\int_{0}^{1} \frac{x ^{2}}{x ^{3} + 1} d x

\frac{d}{d x} (\frac{sin ( \frac{1}{x} ) x ^{3}}{ln ( x )})

a re a 1 - x^{2}, x^{6} - x + 1

x \to - \infty lim (x^{4} + x)