integral of (2^x-3^x)/(5^x)

解

\int \frac{2 ^{x} - 3 ^{x}}{5 ^{x}} d x

\frac{2 ^{x}}{ln ( \frac{2}{5} ) \cdot 5 ^{x}} - \frac{3 ^{x}}{ln ( \frac{3}{5} ) \cdot 5 ^{x}} + C

解答ステップ

\int \frac{2 ^{x} - 3 ^{x}}{5 ^{x}} d x

拡張

\frac{2 ^{x} - 3 ^{x}}{5 ^{x}} : (\frac{2}{5})^{x} - (\frac{3}{5})^{x}

= \int (\frac{2}{5})^{x} - (\frac{3}{5})^{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}{5})^{x} d x - \int (\frac{3}{5})^{x} d x

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

\int (\frac{3}{5})^{x} d x = \frac{3 ^{x}}{ln ( \frac{3}{5} ) \cdot 5 ^{x}}

= \frac{2 ^{x}}{ln ( \frac{2}{5} ) \cdot 5 ^{x}} - \frac{3 ^{x}}{ln ( \frac{3}{5} ) \cdot 5 ^{x}}

定数を解答に追加する

= \frac{2 ^{x}}{ln ( \frac{2}{5} ) \cdot 5 ^{x}} - \frac{3 ^{x}}{ln ( \frac{3}{5} ) \cdot 5 ^{x}} + C