2 x ? 3 x + 1 ? 4 x + 2 ? d x =

Question

$\int 2^{x} 3^{x + 1} 4^{x + 2} d x =$

Smart Achievers · Accepted Answer

(2)

Ans. $\frac{2^{x} 3^{x + 1} 4^{x + 2}}{\log 2 + \log 4 + \log 3} + C$

Explanation

$\int 2^{x} 3^{x + 1} 4^{x + 2} d x = 16 \times 3 \int 2^{x} 3^{x} 4^{x} d x$

$= 48 \int {(24)}^{x} d x = \frac{48 {(24)}^{x}}{\log 24} + C$

$= \frac{2^{x} 3^{x + 1} 4^{x + 2}}{\log 2 + \log 4 + \log 3} + C$

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