Can this Integral be Solved with a Simple Substitution?

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Homework Help Overview

The discussion revolves around the integral \(\int \frac{dx}{\sqrt{x} \left(1+x^2\right)}\) and the attempts to solve it using substitutions and partial fraction decomposition. Participants are exploring methods related to calculus and integration techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts a substitution \(t = \sqrt{x}\) but struggles with the resulting integral. Some participants suggest factoring the denominator and using partial fractions, while others provide hints about polynomial decomposition. There is also a discussion about the validity of certain algebraic manipulations and the complexity of polynomial integrals.

Discussion Status

Participants are actively engaging with the problem, offering various approaches and hints. While some guidance has been provided regarding factoring and partial fractions, there is no explicit consensus on the best method to proceed. The discussion remains open with multiple lines of reasoning being explored.

Contextual Notes

There is an acknowledgment of the challenges posed by polynomial integrals, with some participants noting that certain decompositions may be rare or complex. The original poster's struggle with the integral suggests a need for further clarification on the methods discussed.

twoflower
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Hi,

I'm trying find this integral:

[tex] \int \frac{dx}{\sqrt{x} \left(1+x^2\right)}[/tex]

I tried the substitution [itex]t = \sqrt{x}[/itex] and I got

[tex] 2 \int \frac{dt}{1+t^4}[/tex]

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.
 
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Make your [itex]t= \sqrt{x}[/itex] substitution, then factor the denominator:

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

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

[tex]\int \frac{dt}{1+t^{4}}[/tex]

HINT:

[tex]\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}}[/tex]

Now use that

[tex]a^{2}-b^{2}\equiv (a+b)(a-b)[/tex]

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.

[tex]x = \tan^2{t} \Longrightarrow x^2 = \tan^4{t} = \sin^4{t}\sec^4{t} \neq \sin^2{t}\sec^2{t}[/tex]
 
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


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

[tex]\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[/tex]


[tex]+\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[/tex]

[tex]\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[/tex]

[tex]\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[/tex]
 
Last edited:
Yeah. It's always possible, but not easy in general :smile:
 

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