How to solve this integral question ?

[ssho88]

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

 

[Deleted member 4993]

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

View attachment 33516

Substitute

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

 

[BigBeachBanana]

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

View attachment 33516

The integral does not have an elementary anti-derivative.

 

[ssho88]

Substitute

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

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.

 

[nasi112]

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,

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

 

[Steven G]

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.

Just because this substitution did not work is no reason to assume that the integral doesn't have an elementary solution.

 

[ssho88]

Is there a chance that you mistyped the integral?

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

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

The question is CORRECT, no typo.

 

[ssho88]

Just because this substitution did not work is no reason to assume that the integral doesn't have an elementary solution.

Agree. Could you please show me the solution step by step ?

Hypergeometric functions ? Taylor series ?

 

[nasi112]

Agree. Could you please show me the solution step by step ?

Hypergeometric functions ? Taylor series ?

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


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

 

[ssho88]

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


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

Thanks for your reply.

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

 

[topsquark]

Thanks for your reply.

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

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.)

-Dan

 

[nasi112]

Thanks for your reply.

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

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.