How do we solve (i.e. get the closed form of) $\int \sqrt{1 + t^2} dt$ ?
The Wolfram page shows the closed form of it but not the steps in solving it.
I think I need some algebraic trick..
I thought of some hyperbolic functions and their derivatives and equalities:
If I make $ t = \sinh x$, then I can reduce the integral to $$ \int \cosh ^2 x dx $$ though I'm not sure if this is valid. But from here, I cannot go on evaluating the integral...
Often forgotten in current Calculus textbooks. Are the Euler's substitutions that allow you to solve not only that integral but every one of the form $\int R(\sqrt{ax^2+bx+c},x)\text{d}x$, where $R(x,y)$ is any rational function.
These substitutions transform your integral (and any of the form above) into the integral of a rational function. From there the problem is solved because we know how to compute integrals of any rational function (archived here).
PS: The algorithm I linked for computing integrals of rational functions is not the only one. There are many others.
PSS: This is a problem I have seen asked uncountable times. The blame is clearly not on the students, but on the professors, on the millions expended on several editions of certain Calculus books that do not give mathematical culture, but easy ways to grade students and millions to the authors. It should be common knowledge that all these integrals can be easily solved, and keep the trigonometric substitutions as a trick that is efficient in some cases and only in some cases. We can only advance faster if we don't keep stumbling on the stones that we have ways to avoid.
put $t=\sinh x=\frac{e^x-e^{-x}}{2}$ then you'll obtain $$\int\sqrt{1+t^2}dt=\int\cosh^2 x dx=\int\left(\frac{e^x+e^{-x}}{2}\right)^2dx=\frac{1}{8}(e^{2x}-e^{-2x}+4x)+C=\frac{1}{2}\left(\left(\frac{e^x-e^{-x}}{2}\right)\left(\frac{e^x+e^{-x}}{2}\right)+x\right)+C=\frac{1}{2}\left(t\sqrt{1+t^2}+\sinh^{-1}t\right)+C$$
You wrote that:
The Wolfram page shows the closed form of it but not the steps in solving it.
This is not exactly true. If you click on "show step-by-step solution", then you will see how they got the result - but you need to be registered there. Unless you buy some of their products, you can get 3 such solutions per day. (Of course, they can change their policy in the future.)
They are many similar sites, where you can get calculation of integrals including the way to the solution, for example Mathematical Assistant on Web aka MAW. (Maybe we even have a question about such systems on this site - if not, this might be a reasonable and useful question to ask here.)
I'll put here what WA shown me for your integral:
You could use the defintion $\cosh(x)=\dfrac{\operatorname{exp}(x)+\operatorname{exp}(-x)}2$, square this and integrate term wise.
To solve any hyperbolic trig problem, solve the corresponding trig problem and just adjust each step along the way. So in trig we use $\cos^2=(1-\cos(2x))/2$. Here, we use the corresponding hyperbolic identity, $\cosh^2=(1+\cosh(2x))/2$.
Yes, your reduction is valid. Then note $\cosh x=\frac 12 (\exp(x)+\exp(-x))$, so $\cosh^2 x=\frac 14 (\exp(2x) +2 + \exp(-2x))$. You can probably integrate those terms.
