# Trigonometric substitution

• B
Rhapsody83
Last night I tried to calculate from an automatically generated Wolfram Alpha problem set:
$$\int{\frac{1}{\sqrt{x^2+4}}}dx$$
I answered $$\ln({\frac{\sqrt{x^2+4}}{2}+\frac{x}{2}})+C$$
$$\ln({\sqrt{x^2+4}+x})+C$$
I couldn't see where I had gone wrong, so I tried differentiating my answer in Wolfram Alpha and the answer sheet one as well and both gave
$$\frac{1}{\sqrt{x^2+4}}$$
I'm confused as this suggests that both are correct answers to the question. Can somebody please tell me what is going on as I thought every function had a unique integral? Are both or only one of the answers correct?

PS There should be absolute value brackets around my integral answers but I'm not sure how to do them in LaTeX.

• PeroK

Mentor
2022 Award
You didn't make a mistake. The two expressions are the same with only different constants.
$$\log\left(\dfrac{\sqrt{x^2+4}}{2}+\dfrac{x}{2}\right)+C=\log\left(\dfrac{1}{2}\cdot\left(\sqrt{x^2+4}+x\right)\right)+C=\log\left(\sqrt{x^2+4}+x\right)+\underbrace{\log\left(\dfrac{1}{2}\right)+C}_{=:C'}$$

• • vanhees71, jedishrfu, PeroK and 1 other person
Rhapsody83
Thanks for this, I was very confused, but it makes sense now. Am I right in thinking that one that is not divided by two is the preferred one as it is a little simpler?

Mentor
2022 Award
Thanks for this, I was very confused, but it makes sense now. Am I right in thinking that one that is not divided by two is the preferred one as it is a little simpler?
It is finally a matter of taste. I would prefer the version without an additional factor that is swallowed by the constant anyway. I mean, you could even write ##\log\left(\dfrac{\sqrt{x^2+4}}{\pi}+ \dfrac{x}{\pi}\right),## which, of course, doesn't make much sense. So, yes, put all unnecessary factors in the logarithm (summands outside the logarithm) into ##C.##

This question doesn't come up if you consider definite integrals.

• • vanhees71 and Rhapsody83
Homework Helper
Gold Member
2022 Award
... more generally:
$$\frac d {dx}\bigg ( \ln (\frac{f(x)}{k}) \bigg ) = \frac d {dx}\bigg ( \ln (f(x)) - \ln k \bigg ) = \frac d {dx}\bigg ( \ln (f(x))\bigg )$$Where ##k## is a constant.

• • vanhees71 and Rhapsody83
Rhapsody83
I think I fully understand what is going on now and PeroK's formula, which is obvious when you see it, makes it even clearer.

• vanhees71