# How to solve this integral question ?

Staff member

#### BigBeachBanana

##### Senior Member
• AvgStudent

#### 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,

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

• AvgStudent

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

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

• topsquark

#### 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

• nasi112

#### nasi112

##### Full Member
• 