Integral of $\frac{1}{x\sqrt{4x + 1}}$: Solutions and Explanations

[PhysicsinCalifornia]

$$\int \frac{1}{x\sqrt{4x + 1}}dx$$
I let $$u= \sqrt{4x+1}$$, $$u^2 = 4x+1$$
So $$\frac{1}{2}u du = dx$$

$$\int \frac{\frac{1}{2}u}{(\frac{u^2-1}{4})*((\sqrt{u})^2)}du$$
$$\int \frac{2u}{(u^2 - 1)*(u)} du$$
$$\int \frac{2}{u^2 - 1} du$$
$$2\int \frac{1}{u^2 -1}du$$

Can anyone help me anti-differentiate what the integrand is?

 

[James R]

Isn't it related to the inverse hyperbolic tangent of u?

 

[amcavoy]

Try u=sinθ.

 

[Muzza]

I'd suggest partial fraction decomposition...

 

[TD]

Expanding that fractions gives:

$$\frac{1}<br /> {{u^2 - 1}} = \frac{1}<br /> {{2\left( {u - 1} \right)}} - \frac{1}<br /> {{2\left( {u + 1} \right)}}$$

 

[James R]

Partial fractions work. So does u=tanh t.