Write a
quadratic function with zeroes -6 and -4

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Ответ:

kkeith121p6ujlt

12.08.2021

Mathematics

The solution is $\frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] + C$

Step-by-step explanation:

From the question

The function given is $f(x) = \frac{e^{2x}}{ 25 + e^{4x}} dx$

The indefinite integral is mathematically represented as

$\int\limits {\frac{e^{2x}}{ 25 + e^{4x}}} \, dx$

Now let $e^{2x} = u$

=> $\frac{du}{dx} 2e^{2x}$

=> $2 e^{2x}dx = du$

$\int\limits {\frac{e^{2x}}{ 25 + e^{4x}}} \, dx = \int\limits {\frac{1}{ 2(25 + u^2)} } \, du$

$= \frac{1}{2} \int\limits {\frac{1}{ 25 + u^2)} } \, du$

$= \frac{1}{2} \int\limits {\frac{1}{ 5^2 + u^2)} } \, du$

$= \frac{1}{2} \frac{tan^{-1} [\frac{u}{5} ]}{5} + C$

Now substituting for u

$\frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] + C$

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Write a quadratic function with zeroes -6 and -4