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

The integral of 2^{3x+log_{2}(x)} is (8^xx)/(3ln(2))-(8^x)/(9ln^2(2))+C

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integral of 2^{3x+log_{2}(x)}

\int 2^{3 x + l o g_{2} (x)} d x

\frac{8 ^{x} x}{3 ln ( 2 )} - \frac{8 ^{x}}{9 ln ^{2} ( 2 )} + C

Solution steps

\int 2^{3 x + l o g_{2} (x)} d x

Simplify

2^{3 x + l o g_{2} (x)} : 8^{x} x

= \int 8^{x} x d x

Apply Integration By Parts

= \frac{8 ^{x} x}{3 ln ( 2 )} - \int \frac{8 ^{x}}{3 ln ( 2 )} d x

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

= \frac{8 ^{x} x}{3 ln ( 2 )} - \frac{8 ^{x}}{9 ln ^{2} ( 2 )}

Add a constant to the solution

= \frac{8 ^{x} x}{3 ln ( 2 )} - \frac{8 ^{x}}{9 ln ^{2} ( 2 )} + C

What is the integral of 2^{3x+log_{2}(x)} ?
The integral of 2^{3x+log_{2}(x)} is (8^xx)/(3ln(2))-(8^x)/(9ln^2(2))+C