How do you integrate (a^x + b^x)^3 /((a^x)*(b^x))

[The_ArtofScience]

Homework Statement

Integrate (a^x + b^x)^3 /((a^x)*(b^x))

The Attempt at a Solution

I have only 2 semesters of calculus under my belt yet nothing in my experience has taught me to do anything like this

 

[tiny-tim]

Hi The_ArtofScience!

(try using the X² tag just above the Reply box )

Integrate (a^x + b^x)^3 /((a^x)*(b^x))

erm … just expand it …

(a²/b)^x + …

 

[The_ArtofScience]

The algebra that I did from expanding it like you suggested was:

(a^3x + 3a^2x b^x + 3a^x b^2x + b^3x)/(a^x b^x)
= (a^2 /b)^x + 3a^x + 3b^x + (b^2/a)^x

But then what do I do with this strange exponent? I don't know how to integrate an expression like b^x

 

[Born2bwire]

You can find it easily enough by a search but you can derive it pretty easily too.

\log_b(b^x) = x
\frac{d}{dx}\log_b(b^x) = 1
\frac{d}{dx}\log_b(b^x) = \frac{d}{du} \log_b(u) \cdot \frac{d b^x}{dx} where u = b^x.

Convert log_b(u) = \frac{\ln(u)}{\ln(b)} and go from there.

 

[tiny-tim]

But then what do I do with this strange exponent? I don't know how to integrate an expression like b^x

(please use the X² tag just above the Reply box … it's much easier to read)

Yes, use Born2bwire's method,

or write b^x = (e^ln(b))x = e^ln(b)x,

so (b^x)' = ln(b)b^x

 

[Born2bwire]

Oh sure, do it the easy and efficient way. ;)

 

[The_ArtofScience]

Hi I just read the messages and I appreciate both of your help. Thanks! I get it now. It was silly for me not to see it before I asked