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Q: Using the substitution of $ u = \dfrac{e^{2x}}{5}$, write this integrand as a function of u: $$\int \frac{e^{2x}}{\sqrt{25-e^{4x}}}dx$$

I'm stuck at substituting u into the denominator. The best I can get is: $$\int \frac{5u}{\sqrt{25-{ \frac {5}{2} (\frac{2}{5}e^{4x}})}}du$$

I've tried making it $$\int \frac{5u}{\sqrt{25-{ \frac {5}{2} (\frac{2}{5}e^{2x}})^2}}du$$ but i guess it's wrong too as the items in the bracket will also be squared which is wrong.

Can anyone help me with this? Thank you!

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    $\begingroup$ $e^{4x}=(e^{2x})^{2}=25u^{2}$. Also $dx=\frac 5 {2u} du$. $\endgroup$ – Kabo Murphy May 21 at 6:09
  • $\begingroup$ @KaviRamaMurthy Thank you!!! $\endgroup$ – user672518 May 21 at 6:12
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    $\begingroup$ @KaviRamaMurthy Why are you answering in a comment? $\endgroup$ – Arthur May 21 at 6:14
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When you say $5u=e^{2x}$ then $5du=2e^{2x}dx$ or $\dfrac52du=e^{2x}dx$ so $$\int \frac{e^{2x}dx}{\sqrt{25-e^{4x}}} = \int \dfrac{\dfrac52du}{\sqrt{25-25u^2}}=\dfrac12\int\dfrac{du}{\sqrt{1-u^2}}$$

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  • $\begingroup$ I was just about to write something like this, but then I got a visitor in my office. (+1) $\endgroup$ – mickep May 21 at 7:23
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Hint:

Let $e^{2x}=5\sin\theta$.

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    $\begingroup$ Even though that might be a good substitution to do, it does not answer the question asked, as far as I can see. $\endgroup$ – mickep May 21 at 6:22
  • $\begingroup$ @mickep The question was “Can anyone help me with this?” Giving hints to practice problems from homeworks etc. is standard practice here. Keep challenging yourself, make the above substitution, and see where it takes you.... $\endgroup$ – gen-z ready to perish May 21 at 6:25

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