# Tricky Exponential/Logarithmic Indefinite Integrals

1. Oct 4, 2009

### Liquid7800

Hi,
Our professor has only taught us these methods for Integration...thus far:
Direct Integration
Substituion Method

So theoretically we should be able to solve these problems by substitution, direct integration...(double substitution) & by algebraic manipulation etc.

Ive attached 3 attempts at answers for 3 problems:

This first problem I am curious how our professor got "the answer in question"...Ive tried my best but all I could get was "my answer" the logic seems correct...see image

This second problem I am curious why our professor also gave "the answer in question"...Ive tried my best but I would like to know if I am solving this correctly "my answer"...see image

This last problem I am just curious to see if this is the correct way to solve this problem "my answer"...see image

2. Oct 5, 2009

### HallsofIvy

Staff Emeritus
Your difficulty appears to be bad algebra. In the first problem you have
$$\frac{2e^{2x}- 8/3}{2e^{2x}- 5}$$
and cancel the "$2e^{2x}$" terms. You can't do that.

In the second problem you have, in one equation,
$$\frac{1}{\sqrt{2+ u^3}}[/itex] and in the next, the square root has magically disappeared! In the third problem, you have [tex]\frac{1}{2}\int 9^{u^2} du$$
and somehow that becomes
$$\frac{1}{2}\int log_9(u)du[/itex] How did the exponential become a logarithm? 3. Oct 5, 2009 ### Liquid7800 Thanks for the reply and suggestions... Ive since gotten the answers to problems #1(didnt balance the algebra correctly) and #3 (used the wrong integral formula).. However, I believe #2 is correct...perhaps my posted image is bad...I dont have the equation [tex] \frac{1}{\sqrt{2+ u^3}}$$
anywhere...or at least I shouldn't!

4. Oct 5, 2009

### Dick

In #2 after you do the substitution you have 1/sqrt(1+u^2). But the integration formula you apply is for 1/(a^2+u^2) (without the square root). You need an integration formula for 1/sqrt(u^2+a^2).

5. Oct 5, 2009

### Billy Bob

In problem 2, the obvious and natural method to evaluate $$\int\frac{1}{\sqrt{1+u^2}}\,du$$ is trig substitution. Unfortunately, you said you are not allowed to use that method.

You could use the completely unmotivated trick of multiplying numerator and denominator by $$u+\sqrt{1+u^2}$$. Then the substitution $$w=u+\sqrt{1+u^2}$$ works.