integral of (2^{x+1}-5^{x-1})/(10^x)

解

\int \frac{2 ^{x + 1} - 5 ^{x - 1}}{1 0 ^{x}} d x

- \frac{2}{ln ( 5 ) \cdot 5 ^{x}} + \frac{1}{5 ln ( 2 ) \cdot 2 ^{x}} + C

解答ステップ

\int \frac{2 ^{x + 1} - 5 ^{x - 1}}{1 0 ^{x}} d x

\frac{2 ^{x + 1} - 5 ^{x - 1}}{1 0 ^{x}} = (\frac{2 ^{x + 1}}{1 0 ^{x}} - \frac{5 ^{x - 1}}{1 0 ^{x}})

= \int \frac{2 ^{x + 1}}{1 0 ^{x}} - \frac{5 ^{x - 1}}{1 0 ^{x}} d x

総和規則を適用する:

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

= \int \frac{2 ^{x + 1}}{1 0 ^{x}} d x - \int \frac{5 ^{x - 1}}{1 0 ^{x}} d x

\int \frac{2 ^{x + 1}}{1 0 ^{x}} d x = - \frac{2}{ln ( 5 ) \cdot 5 ^{x}}

\int \frac{5 ^{x - 1}}{1 0 ^{x}} d x = - \frac{1}{5 ln ( 2 ) \cdot 2 ^{x}}

= - \frac{2}{ln ( 5 ) \cdot 5 ^{x}} - (- \frac{1}{5 ln ( 2 ) \cdot 2 ^{x}})

簡素化

= - \frac{2}{ln ( 5 ) \cdot 5 ^{x}} + \frac{1}{5 ln ( 2 ) \cdot 2 ^{x}}

定数を解答に追加する

= - \frac{2}{ln ( 5 ) \cdot 5 ^{x}} + \frac{1}{5 ln ( 2 ) \cdot 2 ^{x}} + C