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Q. 49

Expert-verified
Found in: Page 362

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Use integration formulas to solve each integral in Exercises 21–62. You may have to use algebra, educated guess- and- check, and/or recognize an integrand as the result of a product, quotient, or chain rule calculation. Check each of your answers by differentiating. (Hint for Exercise 54: $\mathrm{tan}\left(x\right)=\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}$).$\int \frac{{e}^{3x}-2{e}^{4x}}{{e}^{2x}}dx$.

The value of the given integral will be ${e}^{x}-{e}^{2x}+c$.

See the step by step solution

## Step 1. Given Information.

Given the integral: $\int \frac{{e}^{3x}-2{e}^{4x}}{{e}^{2x}}dx$.

## Step 2. Formula involved.

$\int {e}^{x}dx={e}^{x}+c.$ and $\int {e}^{ax}dx=\frac{{e}^{ax}}{a}+c,whereaisacons\mathrm{tan}t.$

## Step 3. Solving the integral.

$\int \frac{{e}^{3x}-2{e}^{4x}}{{e}^{2x}}dx\phantom{\rule{0ex}{0ex}}=\int \left({e}^{3x-2x}-2{e}^{4x-2x}\right)dx\phantom{\rule{0ex}{0ex}}=\int \left({e}^{x}-2{e}^{2x}\right)dx\phantom{\rule{0ex}{0ex}}=\int {e}^{x}dx-\int 2{e}^{2x}dx\phantom{\rule{0ex}{0ex}}={e}^{x}-2\frac{{e}^{2x}}{2}+c\phantom{\rule{0ex}{0ex}}={e}^{x}-{e}^{2x}+c.$