Completing Integral problem using substitution: Integral [ ( ln sqr(x) ) / x ] dx

[muzik]

I've used "View an example" on the Pearson website for a Integral problem using substitution.
I understand the process the example had until I got to the end.

The problem is an indefinite integral

Integral [ ( ln sqr(x) ) / x ] dx

The answer came out to

( ln sqr(x) )^2 + C

The part I don't understand is just before the final answer. which is

1/4 ( ln x)^2 + C. How or why does this convert into

( ln sqr(x) )^2 + C

Where did the 1/4 go? and why did it x turn into square root x? ​

During the process in the beginning I had a hard time determining what the substitution should be. Using the example it mentioned use the Log rule.

 

[Dr.Peterson]

I've used "View an example" on the Pearson website for a Integral problem using substitution.
I understand the process the example had until I got to the end.

The problem is an indefinite integral

Integral [ ( ln sqr(x) ) / x ] dx

The answer came out to

( ln sqr(x) )^2 + C

The part I don't understand is just before the final answer. which is

1/4 ( ln x)^2 + C. How or why does this convert into

( ln sqr(x) )^2 + C

Where did the 1/4 go? and why did it x turn into square root x? ​

During the process in the beginning I had a hard time determining what the substitution should be. Using the example it mentioned use the Log rule.

You didn't say what substitution they used; when I did it, I went directly to the final form. If I got 1/4 ( ln x)^2 + C, I'd leave it that way, because it seems simpler!

But as for your question, if I wanted to get sqrt(x) in my answer, I would do this:

1/4 (ln(x))^2 = 1/2^2 (ln(x))^2 = (1/2 ln(x))^2 = (ln(x^{1/2}))^2 = (ln(sqrt(x)))^2
​

Alternatively, if you want to check that the last step equals the next to last, you could do it in reverse, simplifying (ln(sqrt(x)))^2. (What they did is the reverse.)

 

[tkhunny]

I've used "View an example" on the Pearson website for a Integral problem using substitution.
I understand the process the example had until I got to the end.

The problem is an indefinite integral

Integral [ ( ln sqr(x) ) / x ] dx

The answer came out to

( ln sqr(x) )^2 + C

The part I don't understand is just before the final answer. which is

1/4 ( ln x)^2 + C. How or why does this convert into

( ln sqr(x) )^2 + C

Where did the 1/4 go? and why did it x turn into square root x? ​

During the process in the beginning I had a hard time determining what the substitution should be. Using the example it mentioned use the Log rule.

The only other thing that changed was the "sqr(x)". It has to be in there. See if you can find it.

 

[muzik]

You didn't say what substitution they used; when I did it, I went directly to the final form. If I got 1/4 ( ln x)^2 + C, I'd leave it that way, because it seems simpler!

But as for your question, if I wanted to get sqrt(x) in my answer, I would do this:

1/4 (ln(x))^2 = 1/2^2 (ln(x))^2 = (1/2 ln(x))^2 = (ln(x^{1/2}))^2 = (ln(sqrt(x)))^2
​

Alternatively, if you want to check that the last step equals the next to last, you could do it in reverse, simplifying (ln(sqrt(x)))^2. (What they did is the reverse.)

The subtitution used in the example u = ln(x) after using the log rule.
So using the log rule

Integral ( ln sqrt(x) / x ) dx

= Integral ( ln (x)^1/2 / x ) dx , use log rule to move 1/2 in front of ln
so,
= Integral ( 1/2 ln(x) / x ) dx , now use u = ln(x). and so forth..

but back to your example above, I don't understand this portion

1/2^2 (ln(x))^2 = (1/2 ln(x))^2

how does 1/2^2 turn into just 1/2?

Thanks​

 

[muzik]

The only other thing that changed was the "sqr(x)". It has to be in there. See if you can find it.

I see the square root x, in the equation. I just don't understand where the 1/4 went or how it contributes to the square root x.

Or I'm not understanding your comment correctly.

 

[mmm4444bot]

Using properties of exponents and logarithms:

\(\displaystyle \bigg( ln(\sqrt{x}) \bigg)^{2} + \; C\)

\(\displaystyle \bigg( ln(x^{1/2}) \bigg)^{2} + \; C\)

\(\displaystyle \bigg( \frac{1}{2} \cdot ln(x) \bigg)^{2} + \; C\)

\(\displaystyle \frac{1}{4} \cdot ln(x)^{2} + C\)

 

[muzik]

Using properties of exponents and logarithms:

\(\displaystyle \bigg( ln(\sqrt{x}) \bigg)^{2} + \; C\)

\(\displaystyle \bigg( ln(x^{1/2}) \bigg)^{2} + \; C\)

\(\displaystyle \bigg( \frac{1}{2} \cdot ln(x) \bigg)^{2} + \; C\)

\(\displaystyle \frac{1}{4} \cdot ln(x)^{2} + C\)

I see, now I understand.
Thank you very much.

 

[muzik]

Thank you all for you help! I got the help I was looking for.

 

[muzik]

You didn't say what substitution they used; when I did it, I went directly to the final form. If I got 1/4 ( ln x)^2 + C, I'd leave it that way, because it seems simpler!

But as for your question, if I wanted to get sqrt(x) in my answer, I would do this:

1/4 (ln(x))^2 = 1/2^2 (ln(x))^2 = (1/2 ln(x))^2 = (ln(x^{1/2}))^2 = (ln(sqrt(x)))^2
​

Alternatively, if you want to check that the last step equals the next to last, you could do it in reverse, simplifying (ln(sqrt(x)))^2. (What they did is the reverse.)

Now i see what you were trying to show me. Sorry a little hard to understand with all the brackets.
Thank you for you help.