# What is the integral of #int ((x^2-1)/sqrt(2x-1) )dx #?

By signing up, you agree to our Terms of Service and Privacy Policy

To find the integral of (\int \frac{x^2-1}{\sqrt{2x-1}} , dx), we can use the method of substitution. Let (u = 2x-1), then (du = 2 , dx). Rearranging for (dx), we get (dx = \frac{1}{2} du). Substituting these into the integral, we have:

(\int \frac{x^2-1}{\sqrt{2x-1}} , dx = \int \frac{x^2-1}{\sqrt{u}} \cdot \frac{1}{2} , du)

Expanding (x^2-1) and factoring out (\frac{1}{2}), we obtain:

(\frac{1}{2} \int \left(\frac{x^2}{\sqrt{u}} - \frac{1}{\sqrt{u}}\right) , du)

Splitting the integral into two separate integrals, we get:

(\frac{1}{2} \left(\int \frac{x^2}{\sqrt{u}} , du - \int \frac{1}{\sqrt{u}} , du\right))

Now, we integrate each term separately:

For (\int \frac{x^2}{\sqrt{u}} , du), we can rewrite (x^2) in terms of (u). Since (u = 2x-1), we have (x^2 = \frac{u+1}{2}). Substituting this, we get:

(\int \frac{\frac{u+1}{2}}{\sqrt{u}} , du = \frac{1}{2} \int \frac{u+1}{\sqrt{u}} , du)

This integral can be solved using u-substitution.

For (\int \frac{1}{\sqrt{u}} , du), we have a simple integral:

(\int \frac{1}{\sqrt{u}} , du = 2\sqrt{u} + C)

Substituting back (u = 2x-1), we get the final result:

(\frac{1}{2} \left(\int \frac{x^2}{\sqrt{2x-1}} , dx - 2\sqrt{2x-1}\right) + C)

This is the integral of (\int \frac{x^2-1}{\sqrt{2x-1}} , dx).

By signing up, you agree to our Terms of Service and Privacy Policy

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7