First, I am going through my textbook's table of integrals and I would like to find proofs for them, is there a site with proofs of various integrals? This one specifically here I am trying to prove.
$$ \int \frac{\sqrt{a^2 + x^2}}{x}dx = \sqrt{a^2 + x^2} - a\ln\left|\frac{a + \sqrt{a^2 + x^2}}{x}\right| + C $$
Using $a = 1$ as a base, I plugged the following into symbolab and integral-calculator into :
$$ \int \frac{\sqrt{1 + x^2}}{x} dx $$
Now they both use $u$ substitution of:
$$ u = \sqrt{1 + x^2}, du =\frac{\sqrt{1 + x^2}}{x} $$
The derivative they used is wrong? Should it not be:
$$ du = \frac{x}{\sqrt{1 + x^2}} $$
If I can use the inversed derivative then I can easily solve this—using the square, get $\frac{u^2}{u^2 - 1}$, and solve with long division—but this does not seem correct to me. Is there some rule I missed that allows me to use the inverse substituted derivative?
There should be a $dx$ on the RHS of the last term you wrote.
$u = \sqrt{1+x^2} \rightarrow \frac{du}{dx} = \frac{x}{\sqrt{1+x^2}}$
$du = \frac{x \: dx}{\sqrt{1+x^2}}$