Don't know if this is correct (and final step)

1. Mar 4, 2013

iRaid

1. The problem statement, all variables and given/known data
$$\int_0^1 \frac{4ln(3x)}{\sqrt{x}}dx$$

2. Relevant equations

3. 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 :)

$$\int_0^1 \frac{4ln(3x)}{\sqrt{x}}dx=\lim{t \to 0^{+}} \int_0^{t} \frac{4ln(3x)}{\sqrt{x}}dx$$
$$u=4ln3x, v=2\sqrt{x}\\du=\frac{4}{x}dx, dv=\frac{1}{\sqrt{x}}dx$$
$$(4ln3x)(2\sqrt{x})-\int \frac{8\sqrt{x}}{x}dx=8\sqrt{x}(ln3x)-8\int x^{\frac{-1}{2}}dx=8\sqrt{x}(ln3x)-16\sqrt{x}$$
$$\lim_{t \to 0^{+}} (8\sqrt{x}(ln3x)-16\sqrt{x})|_{t}^{1}=\lim_{t \to 0^{+}} 8(ln3)-16-(8\sqrt{t}(ln3t)-16\sqrt{t})$$

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

Thanks.

2. Mar 4, 2013

SammyS

Staff Emeritus
The indefinite integral is correct. (I've checked your work that far.)

What result does Wolfram give you?

3. Mar 4, 2013

iRaid

Wow I'm stupid I just did it again and I'm not sure why wolfram gives me the answer, but my calculator gives me a different answer?? Log=ln right?

4. Mar 4, 2013

SammyS

Staff Emeritus
That's true for Wolfram.

For you calculator, it's probably log = log10 .

5. Mar 4, 2013

iRaid

That is so stupid -.-"