integral of (9^x+81^x)/(1+81^x)

解答

\int \frac{9 ^{x} + 8 1 ^{x}}{1 + 8 1 ^{x}} d x

\frac{arctan ( 9 ^{x} )}{2 ln ( 3 )} + \frac{1}{4 ln ( 3 )} ln ∣ 1 + 8 1^{x} ∣ + C

求解步骤

\int \frac{9 ^{x} + 8 1 ^{x}}{1 + 8 1 ^{x}} d x

乘开

\frac{9 ^{x} + 8 1 ^{x}}{1 + 8 1 ^{x}} : \frac{9 ^{x}}{1 + 8 1 ^{x}} + \frac{8 1 ^{x}}{1 + 8 1 ^{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 \frac{9 ^{x}}{1 + 8 1 ^{x}} d x + \int \frac{8 1 ^{x}}{1 + 8 1 ^{x}} d x

\int \frac{9 ^{x}}{1 + 8 1 ^{x}} d x = \frac{arctan ( 9 ^{x} )}{2 ln ( 3 )}

\int \frac{8 1 ^{x}}{1 + 8 1 ^{x}} d x = \frac{1}{4 ln ( 3 )} ln ∣ 1 + 8 1^{x} ∣

= \frac{arctan ( 9 ^{x} )}{2 ln ( 3 )} + \frac{1}{4 ln ( 3 )} ln ∣ 1 + 8 1^{x} ∣

解答补常数

= \frac{arctan ( 9 ^{x} )}{2 ln ( 3 )} + \frac{1}{4 ln ( 3 )} ln ∣ 1 + 8 1^{x} ∣ + C

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