Another trig substitution question

[vande060]

Homework Statement

sorry wait a few moments for the details, i hit post on accident prematurely

∫ √(1 + x^2)/x dx

The Attempt at a Solution

∫ √(1 + x^2)/x dx

x = tanϑ , dx sec^2ϑ dϑ -π/2 < ϑ < π/2

√(1 + x^2) = secϑ

∫ (secϑ * sec^2ϑ dϑ )/ tanϑ dϑ

after using trig identities

∫cscϑ dϑ + ∫ tanϑsecϑ dϑ

= -Ln I cscϑ + cotϑ I + secϑ + C

= -LnI [√(1 + x^2) +1]/xI + √(1 + x^2) + C

but the back of the book tells me

= LnI [√(1 + x^2) - 1]/xI + √(1 + x^2) + C

am i missing something here

 

[CompuChip]

I'm not quite sure if you mean
\ln\left( i \sqrt{1 + x^2} + 1} / x \right)
or
\ln\left( i \sqrt{1 + x^2} + 1} \right) / x
or
\ln(i) \cdot \left( \sqrt{1 + x^2} + 1} / x \right)

But it might be possible that the answers are actually the same up to a constant.
You could try using the properties of logarithms to rewrite ln(i √(1 + x²) - 1) to ln(i √(1 + x²) - 1) + k
where k is a constant (which you can then absorb in C).

 

[vande060]

I'm not quite sure if you mean
\ln\left( i \sqrt{1 + x^2} + 1} / x \right)
or
\ln\left( i \sqrt{1 + x^2} + 1} \right) / x
or
\ln(i) \cdot \left( \sqrt{1 + x^2} + 1} / x \right)

But it might be possible that the answers are actually the same up to a constant.
You could try using the properties of logarithms to rewrite ln(i √(1 + x²) - 1) to ln(i √(1 + x²) - 1) + k
where k is a constant (which you can then absorb in C).

I am very sorry, there is no i. I is supposed to be the absolute value bar, I forgot the second one.

here is my answer, rewritten.

= -LnI [√(1 + x^2) +1]/xI + √(1 + x^2) + C

I have edited my first post to include the I properly in both my answer and the book's answer

 

[CompuChip]

Ah, thanks for correcting that.
Just out of curiosity, don't you have a vertical bar (|) on your keyboard?

Anyway, how about something like
- \ln \left| \frac{\sqrt{1 + x^2} + 1}{x} \right|
\cdots = \ln \left| \frac{x}{\sqrt{1 + x^2} + 1} \right|
\cdots = \ln \left| \frac{x (\sqrt{1 + x^2} - 1)}{(\sqrt{1 + x^2} + 1)(\sqrt{1 + x^2} - 1)} \right|
\cdots = \ln \left| \frac{\sqrt{1 + x^2} - 1}{x} \right|

I intentionally skipped some steps and justifications (particularly about pulling operations inside the absolute value signs), I will leave those up to you.

 

[vande060]

Ah, thanks for correcting that.
Just out of curiosity, don't you have a vertical bar (|) on your keyboard?

Anyway, how about something like
- \ln \left| \frac{\sqrt{1 + x^2} + 1}{x} \right|
\cdots = \ln \left| \frac{x}{\sqrt{1 + x^2} + 1} \right|
\cdots = \ln \left| \frac{x (\sqrt{1 + x^2} - 1)}{(\sqrt{1 + x^2} + 1)(\sqrt{1 + x^2} - 1)} \right|
\cdots = \ln \left| \frac{\sqrt{1 + x^2} - 1}{x} \right|

I intentionally skipped some steps and justifications (particularly about pulling operations inside the absolute value signs), I will leave those up to you.

i don't see a vertical bar, sorry. I understand the transformation completely

i am sorry to bother with one more question, but why is the final transformation necessary? it does not seem any more simple.

 

[VietDao29]

i don't see a vertical bar, sorry. I understand the transformation completely

i am sorry to bother with one more question, but why is the final transformation necessary? it does not seem any more simple.

No, of course not. What CompuChip is actually trying to show you is that, they are identically the same. Congratulations, you've come to the right result. :)