Evaluate the integral after figuring out the proper method to use

  • Thread starter Thread starter Painguy
  • Start date Start date
  • Tags Tags
    Integral Method
Painguy
Messages
118
Reaction score
0

Homework Statement


∫ ((2t+3)^2)/t^2 dt


Homework Equations





The Attempt at a Solution


I figured that I would use integration by parts. The problem I'm having is that we haven't actually learned integration by parts, only the u substitution method. I went ahead and read the book on the proof and several examples, but its still a bit new to me so I'm not sure how to approach he problem.

u=1/t^2
du=-dt/3t^3

∫dv∫=(2t+3)^2 dt
v=(4t^3)/3 +6t^2 +9t

4t/3 +6+9/t - ∫ ((4t^3)/3 +6t^2 +9t)/t^3 dt

Is there a more straight forward way of solving this?
 
Physics news on Phys.org
Painguy said:

Homework Statement


∫ ((2t+3)^2)/t^2 dt


Homework Equations





The Attempt at a Solution


I figured that I would use integration by parts. The problem I'm having is that we haven't actually learned integration by parts, only the u substitution method. I went ahead and read the book on the proof and several examples, but its still a bit new to me so I'm not sure how to approach he problem.

u=1/t^2
du=-dt/3t^3

∫dv∫=(2t+3)^2 dt
v=(4t^3)/3 +6t^2 +9t

4t/3 +6+9/t - ∫ ((4t^3)/3 +6t^2 +9t)/t^3 dt

Is there a more straight forward way of solving this?

Sure there is. Just multiply the numerator out. Then split it up and integrate.
 
Dick said:
Sure there is. Just multiply the numerator out. Then split it up and integrate.

Oh wow... Thanks for the help. I feel a little silly right now.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top