Integrate (x^3 + 4x)/sqrt(x^2-4) dx

[anderma8]

I'm trying to integrate the following: int (x^3 + 4x)/sqrt(x^2-4) dx

I let u= x^2-4
so du/2 = xdx

granted, the above also gives me x^2=u+4 so, this gives me:

1/2 int u+8 u^-1/2 du

but the professor has:

1/2 int [u^1/2+8 u^-1/2] du

Did I miss the first u^1/2 somewhere? where did it come from?

 

[anderma8]

Never mind... I found my answer:

(u+8)u^-1/2= u^1/2 + 8u^-1/2


Sorry about that! I forgot the parenthesis!

 

[tacman]

It looks like you made a small mistake in your substitution. Instead of letting u = x^2-4, you should let u = sqrt(x^2-4). This is because when you substitute u = x^2-4, you end up with du = 2xdx, which is not the same as xdx as you have in your work. By using the correct substitution, you will get du = (1/2)sqrt(x^2-4)dx, which is exactly what you need to integrate the given function.

So, the correct way to integrate this function would be:

1/2 int (x^3 + 4x)/sqrt(x^2-4) dx = 1/2 int (u^3 + 8u)/(u) du = 1/2 int (u^2 + 8) du = (1/2)(u^3/3 + 8u) + C

= (1/6)(x^2-4)^(3/2) + 4sqrt(x^2-4) + C

I hope this helps clarify where the u^1/2 term came from in your professor's solution. Remember to always double check your substitutions to avoid any mistakes. Good luck with your integration!