Math Help - Ln Integration Problem

1. Ln Integration Problem

$\int \dfrac{(\ln x)^{3}}{2x}dx$

hint?

2. Re: Ln Integration Problem

$u = \ln{x}$

$du = \frac{1}{x} \, dx$

substitute ...

3. Re: Ln Integration Problem

$\int \dfrac{(\ln x)^{3}}{2x}dx$

$u = \ln x$

$du = \dfrac{1}{x} dx$

$\int \dfrac{(u)^{3}}{du}$

Next step?

4. Re: Ln Integration Problem

What? How did that "dx" get into the denominator? And what became of the "2" in the original integral?

5. Re: Ln Integration Problem

Originally Posted by HallsofIvy
What? How did that "dx" get into the denominator? And what became of the "2" in the original integral?
How can I modify the "2" before putting a "du" in the integral?

6. Re: Ln Integration Problem

$\int \frac{(\ln{x})^3}{2x} \, dx = \frac{1}{2} \int (\ln{x})^3 \cdot \frac{1}{x} \, dx$

substitute ...

$\frac{1}{2} \int u^3 \, du$

7. Re: Ln Integration Problem

Originally Posted by Jason76
How can I modify the "2" before putting a "du" in the integral?
Why would you want to "modify" a number? And why did you not answer my questions?
Your original integral was $\int \frac{(ln(x))^3}{2x}dx$ and skeeter suggested that you make the substitution u= ln(x) so that du= (1/x)dx.

So $\int\frac{(ln(x))^3}{2x}dx= \frac{1}{2}\int (ln(x))^3[(1/x)dx]= \frac{1}{2}\int u^3 du$

You should know that no such substitution will "move" dx into the denominator. There simply is no such thing as " $\int \frac{f(u)}{du}$"!

8. Re: Ln Integration Problem

$\int \dfrac{(\ln x)^{3}}{2x}dx$

$u = \ln x$

$\dfrac{1}{2}du = \dfrac{1}{x} dx$ - Some wrong turn is made here. ???

$\frac{1}{2} \int u^3 \, du$

$\dfrac{1}{2} \dfrac{u^{4}}{4} + C$

$\dfrac{u^{4}}{8} + C$

$\dfrac{\ln(x)^{4}}{8} + C$

The book says the answer is $2(\ln x) + C$, though ???

9. Re: Ln Integration Problem

Originally Posted by Jason76
$\int \dfrac{(\ln x)^{3}}{2x}dx$

$u = \ln x$

$\dfrac{1}{2}du = \dfrac{1}{x} dx$ - Some wrong turn is made here. ???
NO! All that is happening is 1/2 is being taken out of the integral as a factor. It is NOT a part of the substitution!

$\frac{1}{2} \int u^3 \, du$

$\dfrac{1}{2} \dfrac{u^{4}}{4} + C$

$\dfrac{u^{4}}{8} + C$

$\dfrac{\ln(x)^{4}}{8} + C$

The book says the answer is $2(\ln x) + C$, though ???
Well your book is clearly wrong. What do you get if you differentiate 2 ln(x) ? You should get 2/x, which is clearly not the same as your integrand...

10. Re: Ln Integration Problem

So your saying my conclusion was correct, not the book's?

Originally Posted by Prove It
NO! All that is happening is 1/2 is being taken out of the integral as a factor. It is NOT a part of the substitution!

Well your book is clearly wrong. What do you get if you differentiate 2 ln(x) ? You should get 2/x, which is clearly not the same as your integrand...

11. Re: Ln Integration Problem

Oh wait, I see what's going on. We would take out the 1/2 from the get go, and then do the substitution.

So the conversions would look like this:

$u = \ln x$

$du = \dfrac{1}{x}$