Expert Answer to "Evaluate the integral. ∫32x+12x+3dx"

Yulia

Answered question

2021-11-01

Evaluate the integral.

\int \frac{3^{2 x + 1}}{2^{x + 3}} d x

krolaniaN

Skilled2021-11-02Added 86 answers

Step 1
Given,

I = \int \frac{3^{2 x + 1}}{2^{x + 3}} d x

Step 2
Re-write the above integral,

I = \int \frac{3^{2 x + 1}}{2^{x + 3}} d x

= \int \frac{3^{2 x} \times 3^{1}}{2^{x} \times 2^{3}} d x

= \frac{3}{8} \int \frac{3^{2 x}}{2^{x}} d x

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

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

Step 3
Formula used:

\int a^{x} d x = \frac{a^{x}}{\ln a} + C

Step 4
Apply the above formula, we get

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

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

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

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