How to solve this integral question ?

ssho88

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

Staff member

ssho88

New member
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.

Last edited:

nasi112

Full Member
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

Elite Member
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

New member
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

New member
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

Full Member
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

New member
$$\displaystyle \sqrt{1+x^4} = 1 + \frac{x^4}{2} - \frac{x^8}{8} + \frac{x^{12}}{16} + ....$$

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

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

topsquark

Senior Member

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