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

• The_ArtofScience
In summary, integrating the expression (a^x + b^x)^3 /((a^x)*(b^x)) can be done by expanding it and then using the logarithmic method or the chain rule to simplify the expression.

## 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

Hi The_ArtofScience! (try using the X2 tag just above the Reply box )
The_ArtofScience said:
Integrate (a^x + b^x)^3 /((a^x)*(b^x))

erm … just expand it …

(a2/b)x + … 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

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.

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

Yes, use Born2bwire's method,

or write bx = (eln(b))x = eln(b)x,

so (bx)' = ln(b)bx Oh sure, do it the easy and efficient way. ;)

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