What is the integral of 3^{2x}(2^{1-x}+1) ?

The integral of 3^{2x}(2^{1-x}+1) is (9^x(2^{1-x}+1))/(2ln(3))+(ln(2))/(ln(3))*(9^x)/(ln(9/2)*2^x)+C

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

\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

Solution steps

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

Apply Integration By Parts

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

= \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}})

Simplify

= \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}}

Add a constant to the solution

= \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

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What is the integral of 3^{2x}(2^{1-x}+1) ?
The integral of 3^{2x}(2^{1-x}+1) is (9^x(2^{1-x}+1))/(2ln(3))+(ln(2))/(ln(3))*(9^x)/(ln(9/2)*2^x)+C