# Difficult integration question

#### Sudharaka

##### Well-known member
MHB Math Helper
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
Hello, flipflop!

$$\displaystyle \int\frac{dx}{e^x + 2}$$

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

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

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

$$\text{Back-substitute: }\:-\tfrac{1}{2}\ln(1 + 2e^{-x}) + C$$

#### checkittwice

##### Member
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.