MHB Difficult integration question

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The discussion focuses on two integration problems posed by a user named Flipflop. For the first integral, a substitution method involving \(u = e^x + 2\) is suggested, leading to a solution of \(-\frac{1}{2}\ln(1 + 2e^{-x}) + C\). The second integral requires a different substitution, \(x = \frac{9}{16}\sec u\), but Flipflop expresses difficulty in solving it. Participants emphasize the importance of allowing time for responses before providing additional hints or solutions. The conversation highlights the collaborative nature of problem-solving in mathematics forums.
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Flipflop's question from Math Help Forum,

Determine the following:

a)\[\int\frac{1}{e^x+2}\,dx\]

b)\[\int\frac{\sqrt{16x^2-9}}{x}\,dx\]

How do I do part (b). I've been stuck at it for hours, even after looking at all the different types of integration formulas.

Hi Flipflop,

For the first one substitute, \(u=e^x+2\). For the second one substitute, \(x=\dfrac{9}{16}\sec u\). Hope you can continue.
 
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Hello, flipflop!

\displaystyle \int\frac{dx}{e^x + 2}
Divide numerator and denominator by e^x:\;\int\frac{e^{-x}\,dx}{1 + 2e^{-x}}

\text{Let }u \,=\,1 + 2e^{-x} \quad\Rightarrow\quad du \,=\,-2e^{-x}dx \quad\Rightarrow\quad e^{-x}dx \,=\,-\tfrac{1}{2}du

\text{Substitute: }\:\int\frac{-\frac{1}{2}du}{u} \;=\;-\tfrac{1}{2}\int\frac{du}{u} \;=\; -\tfrac{1}{2}\ln|u| + C

\text{Back-substitute: }\:-\tfrac{1}{2}\ln(1 + 2e^{-x}) + C
 
MHB Rules said:
14. Do not give premature explanations of hints. When you see that someone is already helping a
member and is waiting for the original poster to give feedback, please don't give more hints or a
full solution. Wait at least 24 hours so that the original poster has a chance to respond.

With this MHB rule in mind, post #2 should not have been created.
 
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