Integrate x√(3x-1) dx

To integrate x√(3x-1) dx, we can use integration by parts. Let's choose u = x and dv = √(3x-1) dx. Then, we have du = dx and v = (2/3)(3x-1)^(3/2).

Now, we can use the integration by parts formula:

∫u dv = uv - ∫v du

Plugging in our values, we get:

∫x√(3x-1) dx = x(2/3)(3x-1)^(3/2) - ∫(2/3)(3x-1)^(3/2) dx

Simplify the first term:

= (2/3)x(3x-1)^(3/2) - (2/3)∫(3x-1)^(3/2) dx

Now, we can integrate the second term using a substitution. Let's let u = 3x-1, then du = 3dx. This gives us:

= (2/3)x(3x-1)^(3/2) - (2/3)∫u^(3/2) * (1/3) du

= (2/3)x(3x-1)^(3/2) - (2/9)∫u^(3/2) du

= (2/3)x(3x-1)^(3/2) - (2/9)(2/5)u^(5/2) + C

= (2/3)x(3x-1)^(3/2) - (4/45)(3x-1)^(5/2) + C

Therefore, the final answer is:

∫x√(3x-1) dx = (2/3)x(3x-1)^(3/2) - (4/45)(3x-1)^(5/2) + C

Let $$u=3x-1$$ then $$du=3dx$$

Substituting into the integral:

$$\int x\sqrt{3x-1}dx = \int \frac{1}{3}u^{1/2}du$$

$$=\frac{1}{3}\frac{u^{3/2}}{3/2}+C=\frac{1}{9}(3x-1)^{3/2}+C$$