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Difficult integration question

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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.
 
Last edited:

soroban

Well-known member
Feb 2, 2012
409
Hello, flipflop!

[tex]\displaystyle \int\frac{dx}{e^x + 2}[/tex]

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

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

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

[tex]\text{Back-substitute: }\:-\tfrac{1}{2}\ln(1 + 2e^{-x}) + C [/tex]
 

checkittwice

Member
Apr 3, 2012
37
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.