I'd like some clarification on integrating f(ax+b)

Astronomer_X

New member
Hi! Today in class we were learning how to integrate f(ax+b). Most of the exercise has been fine, and I understand all the standard function rules from differentiation. However:

My text book outlined this method of integrating a function in the form of (ax+b)n (Look at Question B, the second one explained):

So far, I understood the premise well; it's similar to the basic integration I learnt last year, but we do not divide by (n+1) after raising the power. Fine. I then tried to apply it to this question:

As you can see, I followed the same logic to be applied. However, this is what the solution/answer actually is:

I can see that they expanded the equation first, but should the same method as outlined in the example not work as well? If so, why/why not? Because imagine if the question I failed at was raised to the power of 3, or 4 or 5- expanding that first would be ridiculously time consuming. Did I just not apply the chain rule right?

MarkFL

Super Moderator
Staff member
I would say the reason you didn't get the correct answer is because the derivative of the "inner" function is not simply a constant. You were trying to apply a method for:

$$\displaystyle f(x)=(ax+b)^n$$

to a function of the form:

$$\displaystyle g(x)=\left(ce^x+d\right)^n$$

This causes an issue with having the correct differential in your integral when you attempt to substitute.

Astronomer_X

New member
I would say the reason you didn't get the correct answer is because the derivative of the "inner" function is not simply a constant. You were trying to apply a method for:

$$\displaystyle f(x)=(ax+b)^n$$

to a function of the form:

$$\displaystyle g(x)=\left(ce^x+d\right)^n$$

This causes an issue with having the correct differential in your integral when you attempt to substitute.
So could I ask how I would apply it to g(x) in your example? Or is that not possible for that type of function?

MarkFL

Super Moderator
Staff member
So could I ask how I would apply it to g(x) in your example? Or is that not possible for that type of function?
The method you've been shown would not work for $$g(x)$$. It would work for this function:

$$\displaystyle h(x)=ae^{bx}\left(ce^{bx}+d\right)^n$$

Here, we could let:

$$\displaystyle y=\left(ce^{bx}+d\right)^{n+1}\implies dy=(n+1)bce^{bx}\left(ce^{bx}+d\right)^n\,dx$$

Do you see now that the differential $$dy$$ is just some constant times $$h(x)\,dx$$?

Astronomer_X

New member
I see. So, I should just go about a g(x) type question by expanding first?

MarkFL

Super Moderator
Staff member
I see. So, I should just go about a g(x) type question by expanding first?
Yes, and hopefully you won't be given a problem of that type where $$n$$ is large, although in theory, the binomial theorem could be used for the expansion.