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

## Homework Statement

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

## 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.

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SammyS
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Homework Helper
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## Homework Statement

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

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

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
Staff Emeritus
Homework Helper
Gold Member
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 .

That's true for Wolfram.

For you calculator, it's probably log = log10 .
That is so stupid -.-"