Can this Integral be Solved with a Simple Substitution?

  • Thread starter Thread starter twoflower
  • Start date Start date
  • Tags Tags
    Integral
AI Thread Summary
The integral in question is ∫ dx/(√x(1+x²), which was initially approached using the substitution t = √x, leading to the transformed integral 2∫ dt/(1+t⁴). The discussion highlights the need for partial fraction decomposition to solve the integral of 1/(1+t⁴), suggesting a specific factoring method for the denominator. Participants emphasize the importance of recognizing polynomial factors and applying partial fractions, although they note that such decompositions can be complex. Ultimately, the solution involves evaluating four simpler integrals, including those related to arctangent and natural logarithm functions.
twoflower
Messages
363
Reaction score
0
Hi,

I'm trying find this integral:

<br /> \int \frac{dx}{\sqrt{x} \left(1+x^2\right)}<br />

I tried the substitution t = \sqrt{x} and I got

<br /> 2 \int \frac{dt}{1+t^4}<br />

which I'm not able to solve. I know it should be some simple substitution, but I can't see any...

Thank you for any help.
 
Physics news on Phys.org
Make your t= \sqrt{x} substitution, then factor the denominator:

1+t^4 = \left(t^2 + \frac{2}{\sqrt{2}}t + 1\right)\left(t^2 - \frac{2}{\sqrt{2}}t + 1\right)

then use partial fractions.
 
Last edited:
U need partial fraction decomposition for the integral

\int \frac{dt}{1+t^{4}}

HINT:

\frac{1}{1+t^{4}}=\frac{1}{1+2t^{2}+t^{4}-2t^{2}}=\frac{1}{\left(1+t^{2}\right)^{2}-(x\sqrt{2})^{2}}

Now use that

a^{2}-b^{2}\equiv (a+b)(a-b)

to get the partial fraction decomposition.

In the end,u'll have to evaluate 4 simple integrals,2 involving artangent & 2 involving natural logarithm.

Daniel.
 
To add to what dexter and I have already said, remember that you can always factor a polynomial over the reals into quadractic and linear factors and apply partial fractions if necessary.
 
That is wrong, though.

x = \tan^2{t} \Longrightarrow x^2 = \tan^4{t} = \sin^4{t}\sec^4{t} \neq \sin^2{t}\sec^2{t}
 
What Data said it's true,but it's very rare that this decomposition/factoring is possible.The polyinomial that u had was really simple,but consider this one


<br /> <br /> \int \frac{dx}{3x^7+2x^6-3x^5+2x^3+8}=

\sum_{R=\rho }R\ln \left( x+\frac{50537\,71425\,23884\,37683\,12969\,72723\,32288}{1\,47925\,11206\,35366\,91747\,99713}R^6+\frac{718\,89436\,36041\,51477\,11255\,98547\,18976}{1\,47925\,11206\,35366\,91747\,99713}R^5


+\frac{12\,11818\,20513\,41769\,59090\,66072\,73984}{1\,47925\,11206\,35366\,91747\,99713}R^4+\frac{53395\,47640\,47920\,70339\,42785\,91360}{1\,47925\,11206\,35366\,91747\,99713}R^3+\frac{28\,14374\,51428\,61007\,71200\,20096}{1\,47925\,11206\,35366\,91747\,99713}R^2

\left +\frac{86\,45544\,28717\,76847\,41993\,27736}{1\,47925\,11206\,35366\,91747\,99713}R+\frac{48554\,15892\,19677\,88753\,18529}{4\,43775\,33619\,06100\,75243\,99139} \right) +C

\mbox{where}\ \rho \ \mbox {is a root of} \ 15418\,67541\,62688 x^7+6\,15789\,52704x^5 \\ +922226688x^4-196992x^3+257184x^2-1520x-243
 
Last edited:
Yeah. It's always possible, but not easy in general :smile:
 
Back
Top