Answer:
![7\text{Ln}\left(e^(2x)+10\right)+C](https://img.qammunity.org/2023/formulas/mathematics/college/idum32s61sortt5mwp0g7vafouu9wnlfe4.png)
This is the same as writing 7*Ln( e^(2x) + 10) + C
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Step-by-step explanation:
Start with the equation
![u = e^(2x)+10](https://img.qammunity.org/2023/formulas/mathematics/college/vb8zx204fwif0h9wiqr7bw9e7jugmoehfi.png)
Apply the derivative and multiply both sides by 7 like so
![u = e^(2x)+10\\\\(du)/(dx) = 2e^(2x)\\\\7(du)/(dx) = 7*2e^(2x)\\\\7(du)/(dx) = 14e^(2x)\\\\7du = 14e^(2x)dx\\\\](https://img.qammunity.org/2023/formulas/mathematics/college/dnpv2ev3uowfevgh3b4u6xjjvv8lyx0j6c.png)
The "multiply both sides by 7" operation was done to turn the 2e^(2x) into 14e^(2x)
This way we can do the following substitutions:
![\displaystyle \int (14e^(2x))/(e^(2x)+10)dx\\\\\\\displaystyle \int (1)/(e^(2x)+10)14e^(2x)dx\\\\\\\displaystyle \int (1)/(u)7du\\\\\\\displaystyle 7\int (1)/(u)du\\\\\\](https://img.qammunity.org/2023/formulas/mathematics/college/ingkizbdqcl1kmuh3w4zued9w0h87jzjcz.png)
Integrating leads to
![\displaystyle 7\int (1)/(u)du\\\\\\7\text{Ln}\left(u\right)+C\\\\\\7\text{Ln}\left(e^(2x)+10\right)+C\\\\\\](https://img.qammunity.org/2023/formulas/mathematics/college/6f9npwn42zdyovd3a4hg4hgulgvc3szxgo.png)
Be sure to replace 'u' with e^(2x)+10 since it's likely your teacher wants a function in terms of x. Also, do not forget to have the plus C at the end. This is a common mistake many students forget to do.
To verify the answer, you can apply the derivative to it and you should get back to the original integrand of
![(14e^(2x))/(e^(2x)+10)](https://img.qammunity.org/2023/formulas/mathematics/college/gb7bql5x9yxxuyjncd40uhw0ynhst46wyh.png)