Step-by-Step Guide to Solving the Integral of (1)/(x sqrt(4x^4-9))

[Cepeda09]

Homework Statement

What is the integral of (1)/(x sqrt(4x^4-9)) I need the steps. Thanks.

Homework Equations

The Attempt at a Solution

 

[jgens]

What have you tried? No one here can give you any help until you've shown some attempt at the solution.

 

[Cepeda09]

I first simplified it to the int of (1)/(x)(2x^2-3)
I also think that I have to use this, int (1)/(1+x^2) = arctan x +c, in order to solve it. But I can't figure it out.

 

[jgens]

I first simplified it to the int of (1)/(x)(2x^2-3)
I also think that I have to use this, int (1)/(1+x^2) = arctan x +c, in order to solve it. But I can't figure it out.

How did you simplify it like that?

I tackled this problem using a substitution, so I can help you if you're willing to work with that. There might be other ways, but I can't figure them out off the top of my head.

 

[pongo38]

When you simplify something like (1)/(x sqrt(4x^4-9)) to (1)/(x)(2x^2-3), you should check it by putting in some value of x, say 1, to see if you have made an error in simplification. In this case what do you find when you do that?

 

[Cepeda09]

I guess I did make a mistake because I am not getting the same thing. What do I need to do to solve it?

 

[Mentallic]

I first simplified it to the int of (1)/(x)(2x^2-3)
I also think that I have to use this, int (1)/(1+x^2) = arctan x +c, in order to solve it. But I can't figure it out.

A common mistake, but you need to lose this bad habit.

$$\sqrt{a^2+b^2}\neq a+b$$ because $$(a+b)^2=a^2+b^2+2ab\neq a^2+b^2$$

So you can't simplify the surd as you did.

 

[jgens]

I guess I did make a mistake because I am not getting the same thing. What do I need to do to solve it?

I solved it with a substitution. What substitution can you use to eliminate the squareroot in the denominator?

 

[pongo38]

Whatever the substitution is, it could be based on the trig equation cos^2 x + sin^2 x = 1, or a similar thing. When this is rearranged as say cos^2 x = 1 - sin^2 x, you get something rather like your denominator. Can you work it from there?