Integration question

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?

Last edited:

James R

Homework Helper
Gold Member
Isn't it related to the inverse hyperbolic tangent of u?

Try u=sinθ.

Muzza

I'd suggest partial fraction decomposition...

TD

Homework Helper
Expanding that fractions gives:

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

James R

Homework Helper
Gold Member
Partial fractions work. So does u=tanh t.

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