# just a test

• Oct 3rd 2008, 01:19 PM
harold
just a test
$
\int \frac {dx}{x \ln x^2} dx$

1) $= \int \frac {dx}{2x \ln x} dx$ This step follows from the property of logs (exponents)

2) $= \frac {1}{2}\int \frac {dx}{x \ln x} dx$ Extract any constants and place in front of the integral

Let $u = \ln x$, then $du = \frac {1}{x} dx.$

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

4) $= \frac {1}{2}\int \frac {1}{u} du$ U-substitution applied

5) $= \frac {1}{2} \ln {\color{black}|}u{\color{black}|} + C$ By Formula C, and $u > 0$

6) $= \frac {1}{2} \ln {\color{black}|}(\ln x){\color{black}|} + C$ By Reverse substitution
• Oct 3rd 2008, 02:04 PM
Jhevon
Quote:

Originally Posted by harold
$
\int \frac {dx}{x ln x^2} dx$

1) $= \int \frac {dx}{2x ln x} dx$ This step follows from the property of logs (exponents)

2) $= \frac {1}{2}\int \frac {dx}{x ln x} dx$ Extract any constants and place in front of the integral

Let $u = ln x$, then $du = \frac {1}{x} dx.$

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

4) $= \frac {1}{2}\int \frac {1}{u} du$ U-substitution applied

5) $= \frac {1}{2} ln u + C$ By Formula C, and $u > 0$

6) $= \frac {1}{2} ln (ln x) + C$ By Reverse substitution

i know this is just a LaTeX test, but just to be a pain.

type \ln x, instead of just ln x, it looks nicer. you get $\ln x$ as opposed to $ln x$, see?

also, $\ln (x^2) = 2 \ln {\color{red}|}x {\color{red}|}$ ....yes, the absolute valus matter

and you also should have absolute values in your solutions, so it is $\frac 12 \ln {\color{red}|} \ln x {\color{red}|} + C$

other than that, good job!
• Oct 4th 2008, 11:13 AM
harold
Thanks Jhevon!! Was wondering why it didn't have the look it should. Good catch with the absolute value signs too!