- #1

- 559

- 8

## Homework Statement

[tex]\int_0^1 \frac{4ln(3x)}{\sqrt{x}}dx[/tex]

## Homework Equations

## The Attempt at a Solution

I really dont want to type all this out, but since I hate when people don't, I will. I hope someone takes the time to help me out :)

[tex]\int_0^1 \frac{4ln(3x)}{\sqrt{x}}dx=\lim{t \to 0^{+}} \int_0^{t} \frac{4ln(3x)}{\sqrt{x}}dx[/tex]

[tex]u=4ln3x, v=2\sqrt{x}\\du=\frac{4}{x}dx, dv=\frac{1}{\sqrt{x}}dx[/tex]

[tex](4ln3x)(2\sqrt{x})-\int \frac{8\sqrt{x}}{x}dx=8\sqrt{x}(ln3x)-8\int x^{\frac{-1}{2}}dx=8\sqrt{x}(ln3x)-16\sqrt{x}[/tex]

[tex]\lim_{t \to 0^{+}} (8\sqrt{x}(ln3x)-16\sqrt{x})|_{t}^{1}=\lim_{t \to 0^{+}} 8(ln3)-16-(8\sqrt{t}(ln3t)-16\sqrt{t})[/tex]

Now I think that just equals 8ln3-16, but I'm not sure and wolfram gives me a different answer..

Thanks.