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- Thread starter ssho88
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SubstituteHow to solve this integral question ? Integral [ ( 1/x^2 * (1+x^4)^0.5) ] dx ?

View attachment 33516

x^2 = tan(u)............continue

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The integral does not have an elementary anti-derivative.How to solve this integral question ? Integral [ ( 1/x^2 * (1+x^4)^0.5) ] dx ?

View attachment 33516

I don't think so ! Could you please show me the solution step by step ?Substitute

x^2 = tan(u)............continue

With substitution x^2 = tan(y), I get :-

(2x dx ) =sec^2(y)dy, and x = sqrt(tan(y))

dx = (1/2) * (1/x) * sec^2(y)dy

(1/x^2)(1+x^4)^0.5 dx = (1/tan(y)) * sqrt(1 + tan^2(y)) * (1/2) * (1/x) * sec^2(y)dy

= (1/tan(y)) * sqrt(1 + tan^2(y)) * (1/2) * (1/sqrt(tan(y))) * sec^2(y)dy

= (1/tan(y)) * sec(y) * (1/2) * (1/sqrt(tan(y))) * sec^2(y)dy

= (1/tan(y)) * sec^3(y) * (1/2) * (1/sqrt(tan(y))) dy

= ????

It does not have an elementary anti-derivative.

Last edited:

How to solve this integral question ? Integral [ ( 1/x^2 * (1+x^4)^0.5) ] dx ?

View attachment 33516

Is there a chance that you mistyped the integral?

If you have to solve this integral with elementary anti-derivative, the integral should be,

\(\displaystyle \int \frac{\sqrt{1 + x^4}}{x^3} \ dx\)

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Just because this substitution did not work is no reason to assume that the integral doesn't have an elementary solution.I don't think so ! Could you please show me the solution step by step ?

With substitution x^2 = tan(y), I get :-

(2x dx ) =sec^2(y)dy, and x = sqrt(tan(y))

dx = (1/2) * (1/x) * sec^2(y)dy

(1/x^2)(1+x^4)^0.5 dx = (1/tan(y)) * sqrt(1 + tan^2(y)) * (1/2) * (1/x) * sec^2(y)dy

= (1/tan(y)) * sqrt(1 + tan^2(y)) * (1/2) * (1/sqrt(tan(y))) * sec^2(y)dy

= (1/tan(y)) * sec(y) * (1/2) * (1/sqrt(tan(y))) * sec^2(y)dy

= (1/tan(y)) * sec^3(y) * (1/2) * (1/sqrt(tan(y))) dy

= ????

It does not have an elementary anti-derivative.

The question is CORRECT, no typo.Is there a chance that you mistyped the integral?

If you have to solve this integral with elementary anti-derivative, the integral should be,

\(\displaystyle \int \frac{\sqrt{1 + x^4}}{x^3} \ dx\)

Agree. Could you please show me the solution step by step ?Just because this substitution did not work is no reason to assume that the integral doesn't have an elementary solution.

Hypergeometric functions ? Taylor series ?

\(\displaystyle \sqrt{1+x^4} = 1 + \frac{x^4}{2} - \frac{x^8}{8} + \frac{x^{12}}{16} + ....\)Agree. Could you please show me the solution step by step ?

Hypergeometric functions ? Taylor series ?

\(\displaystyle \int \frac{\sqrt{1+x^4}}{x^2} \ dx = \int \frac{1}{x^2} + \frac{x^2}{2} - \frac{x^6}{8} + \frac{x^{10}}{16} + .... \ dx = -\frac{1}{x} + \frac{x^3}{6} - \frac{x^7}{56} + \frac{x^{11}}{176} + ....\)

Thanks for your reply.\(\displaystyle \sqrt{1+x^4} = 1 + \frac{x^4}{2} - \frac{x^8}{8} + \frac{x^{12}}{16} + ....\)

\(\displaystyle \int \frac{\sqrt{1+x^4}}{x^2} \ dx = \int \frac{1}{x^2} + \frac{x^2}{2} - \frac{x^6}{8} + \frac{x^{10}}{16} + .... \ dx = -\frac{1}{x} + \frac{x^3}{6} - \frac{x^7}{56} + \frac{x^{11}}{176} + ....\)

I agree with your solution, but this is only approximate solution, is there an exact solution?

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No, Hypergeometric functions (or however you want to express this) do not have exact expressions. We have said this about this integral before. (And on a couple of different Forums.)Thanks for your reply.

I agree with your solution, but this is only approximate solution, is there an exact solution?

-Dan

Only integrals that can be solved with an elementary anti-derivative can have an exact solution. In the other hand, solutions with Hypergeometric functions, Taylor series, or any similar solutions can have only approximations. But these approximations go toward the exact solution as you go toward infinite.Thanks for your reply.

I agree with your solution, but this is only approximate solution, is there an exact solution?