Tricky Exponential/Logarithmic Indefinite Integrals

[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

 

[HallsofIvy]

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]<br /> and in the next, the square root has magically disappeared!<br /> <br /> In the third problem, you have <br /> \frac{1}{2}\int 9^{u^2} du<br /> and somehow that becomes <br /> \frac{1}{2}\int log_9(u)du[/itex]&lt;br /&gt; How did the exponential become a logarithm?

 

[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 don't have the equation
<br /> \frac{1}{\sqrt{2+ u^3}}<br />
anywhere...or at least I shouldn't!

 

[Dick]

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 don't have the equation
<br /> \frac{1}{\sqrt{2+ u^3}}<br />
anywhere...or at least I shouldn't!

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).

 

[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.