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

[iRaid]

Homework Statement

\int_0^1 \frac{4ln(3x)}{\sqrt{x}}dx

Homework Equations

The Attempt at a Solution

I really don't 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.

 

[SammyS]

Homework Statement

\displaystyle \int_0^1 \frac{4\ln(3x)}{\sqrt{x}}dx

Homework Equations

The Attempt at a Solution

I really don't 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{4\ln(3x)}{\sqrt{x}}dx=\lim_{t \to 0^{+}} \int_0^{t} \frac{4\ln(3x)}{\sqrt{x}}dx<br /><br /> u=4\ln3x, v=2\sqrt{x}\\du=\frac{4}{x}dx, dv=\frac{1}{\sqrt{x}}dx<br /><br /> (4\ln3x)(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}<br /><br /> \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.

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

What result does Wolfram give you?

 

[iRaid]

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

What result does Wolfram give you?

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?

 

[SammyS]

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?

That's true for Wolfram.

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

 

[iRaid]

That's true for Wolfram.

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

That is so stupid -.-"