How Do You Simplify Complex Root Integrals?

  • Thread starter Thread starter transgalactic
  • Start date Start date
  • Tags Tags
    Integral Root
transgalactic
Messages
1,386
Reaction score
0
i tried this:
<br /> \int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}=\int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}*\frac{1+\sqrt{x+1}}{1+\sqrt{x+1}}*\frac{1-\sqrt[3]{x+1}+(x+1)^{\frac{3}{2}}}{1-\sqrt[3]{x+1}+(x+1)^{\frac{3}{2}}}<br />
but when i got read of 2 roots i got another two roots which are more complicated
??
 
Physics news on Phys.org
transgalactic said:
<br /> \int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}

Hi transgalactic! :smile:

Hint: substitute! :wink:
 
i tried t=(x+1)^1/3
but it creates anoted roots in the dt
??
 
transgalactic said:
i tried t=(x+1)^1/3
but it creates anoted roots in the dt
??

uhh? :confused:

dx = … ?

anyway, (x+1)1/6 might be easier.
 
but i don't have members of 1/6 power
??
 
Sure you do. a^(1/2) = a^(1/6)^3, and b^(1/3) = b^(1/6)^2.
 
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