Evaluating $\int \sqrt{1 + t^2} dt$?

Solution 1:

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.

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.

Solution 2:

put $t=\sinh x=\frac{e^x-e^{-x}}{2}$ then you'll obtain $$\int\sqrt{1+t^2}dt=\int\cosh^2 x dx=\int\left(\frac{e^x+e^{-x}}{2}\right)^2dx=\frac{1}{8}(e^{2x}-e^{-2x}+4x)+C=\frac{1}{2}\left(\left(\frac{e^x-e^{-x}}{2}\right)\left(\frac{e^x+e^{-x}}{2}\right)+x\right)+C=\frac{1}{2}\left(t\sqrt{1+t^2}+\sinh^{-1}t\right)+C$$