# 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 -.-"