Hey everyone, I am trying to solve the following indefinite integral:

[math]\int \frac{dx}{x\sqrt{4x^2-9}} [/math]

 

And I have got 2Arcsecx+c... but I am afraid it could be wrong... can anyone please check it...

I simply plugged it into mathematica and got

 

[math]{\frac{-\arctan ({\frac{3}

{{\sqrt{-9 + 4\,{x^2}}}}})}{

3}}[/math]

 

Maybe you can spot this by differentianting arctan.

[math]\int \frac {1}{x} \frac {1}{\sqrt {(2x)^2 - 3^2}} dx[/math]

 

Integration by Parts:

 

[math]\int u dv = uv - \int v du[/math]

 

Choose u by LIATE, Logs, Inverse Trigs, Algebraic, Trigs, Exponentials.

 

So in this, we only have algebraic...I would prefer to integrate the [math]\frac {1}{\sqrt {(2x)^2 - 3^2}}[/math].

 

So lets make [math]dv = \frac {1}{\sqrt {(2x)^2 - 3^2}} dx[/math] and [math]u= \frac {1}{x}[/math].

 

Now just use a lil trig sub for dv and your home free, Good Luck.

 

O yea, Have you learned trig sub, maybe not, because if you had you would recognise that your integrals form is of sec substitution, which will always get your integral an arctan somewhere in it. Need help ask.

Substitute, x= 1/z. Then, apart form the constants, the integral becomes of the form 1/sqrt(4/9-z^2).

The final answer should be (1/3).arcCos{3/(2x)} +C [or, -(1/3).arcSin{3/(2x)} +C].

Hey everyone, I am trying to solve the following indefinite integral:

[math]\int \frac{dx}{x\sqrt{4x^2-9}} [/math]

The faster way to solve this, it's by settin' [math]u^2=4x^2-9[/math].

 

The rest follows.