StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q. 49

Expert-verifiedFound in: Page 362

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 ${e}^{x}-{e}^{2x}+c$.

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

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

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

94% of StudySmarter users get better grades.

Sign up for free