integral of 3^{2x}(2^{1-x}+1)

解

\int 3^{2 x} (2^{1 - x} + 1) d x

\frac{9 ^{x} ( 2 ^{1 - x} + 1 )}{2 ln ( 3 )} + \frac{ln ( 2 )}{ln ( 3 )} \cdot \frac{9 ^{x}}{ln ( \frac{9}{2} ) \cdot 2 ^{x}} + C

解答ステップ

\int 3^{2 x} (2^{1 - x} + 1) d x

部分積分を適用する

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

\int - \frac{ln ( 2 ) \cdot 9 ^{x}}{ln ( 3 ) \cdot 2 ^{x}} d x = - \frac{ln ( 2 )}{ln ( 3 )} \cdot \frac{9 ^{x}}{ln ( \frac{9}{2} ) \cdot 2 ^{x}}

= \frac{9 ^{x} ( 2 ^{1 - x} + 1 )}{2 ln ( 3 )} - (- \frac{ln ( 2 )}{ln ( 3 )} \cdot \frac{9 ^{x}}{ln ( \frac{9}{2} ) \cdot 2 ^{x}})

簡素化

= \frac{9 ^{x} ( 2 ^{1 - x} + 1 )}{2 ln ( 3 )} + \frac{ln ( 2 )}{ln ( 3 )} \cdot \frac{9 ^{x}}{ln ( \frac{9}{2} ) \cdot 2 ^{x}}

定数を解答に追加する

= \frac{9 ^{x} ( 2 ^{1 - x} + 1 )}{2 ln ( 3 )} + \frac{ln ( 2 )}{ln ( 3 )} \cdot \frac{9 ^{x}}{ln ( \frac{9}{2} ) \cdot 2 ^{x}} + C