$\int \sqrt (1 + x^2) dx$
let $x = \tan \theta $
then $dx = \sec ^2\theta d\theta $
we have the integral is then: $\int \sec ^3\theta d\theta $
recall: $\tan ^2\theta + 1 = \sec ^2\theta $ and write as: $\int \sec \theta (\sec ^2\theta )d\theta $
continue with integration by parts, by letting:
$u = \sec \theta $
and
$dv = \sec ^2\theta d\theta $
we have:
$v = \tan \theta $
and
$du = \sec \theta \tan \theta d\theta $
and thus,
$uv - \int vdu$
is then:
$\sec \theta \tan \theta - \int \sec \theta \tan ^2\theta d\theta $
recall:
$\tan ^2\theta + 1 = \sec ^2\theta $
we have:
$\sec \theta \tan \theta - \int \sec \theta (\sec ^2\theta - 1) d\theta $
after distribution, altogether we have:
$\int \sec ^3\theta d\theta = \sec \theta \tan \theta - \int \sec ^3\theta d\theta - \int \sec \theta d\theta $
rearranging and collecting like-terms:
$2\int \sec ^3\theta d\theta = \sec \theta \tan \theta - \int \sec \theta d\theta $
$2\int \sec ^3\theta d\theta = \sec \theta \tan \theta - \ln |\sec \theta + \tan \theta |$
$\int \sec ^3\theta d\theta = (1/2)[\sec \theta \tan \theta - \ln |\sec \theta + \tan \theta |]$
+++++++++++++++++++++++++++
without u\sin g integration tables, we can derive
$\int \sec \theta d\theta $
..........
$=\int 1/\cos \theta d\theta $
$=\int \cos \theta /\cos ^2\theta d\theta $
$=\int \cos \theta /(1 - \sin ^2\theta ) d\theta $
let $u = \sin \theta $
then $du = \cos \theta d\theta $
we have: $\int 1/(1 - u^2) du$
partial fractions:
$A/(1 - u) + B/(1 + u) = 1$
$A + B = 1$
$A - B = 0$
$A = 1/2$
$B = 1/2$
we have:
$-(1/2)\ln |1 - u| + (1/2)\ln |1 + u|$
by rules of logarithms:
$(1/2)\ln |(1 + u)/(1 - u)|$
$(1/2)\ln |(1 - u^2)/(1 - u)^2|$
by rules of logarithms:
$\ln |\sqrt (1 - u^2)/(1 - u)|$
recall:
$u = \sin \theta $
$\ln |\sqrt (1 - \sin ^2\theta )/(1 - \sin \theta )|$
$\ln |\sqrt \cos ^2\theta /(1 - \sin \theta )|$
$\ln |\cos \theta /(1 - \sin \theta )|$
$\ln |\cos \theta (1 + \sin \theta )/(1 - \sin ^2\theta )|$
$\ln |\cos \theta (1 + \sin \theta )/\cos ^2\theta |$
$\ln |1/\cos \theta + \sin \theta /\cos \theta |$
$\ln |\sec \theta + \tan \theta |$
++++++++++++++++++++++++++++
but the integral of $\int \sec ^3\theta d\theta $ is:
$\int \sec ^3\theta d\theta = (1/2)[\sec \theta \tan \theta - \ln |\sec \theta + \tan \theta |]$ + CONS\tan T
and \sin ce $\tan \theta = x = x/1 =$ opposite/adjacent, with the pythagorean theorem, we derive: hypotenuse = $\sqrt (1 + x^2)$ and thus, $\sec \theta $ = hypotenuse/adjacent = $\sqrt (1 + x^2)$
$\int \sqrt (1 + x^2) dx$ $= (1/2)[x\sqrt (1 + x^2) - \ln |\sqrt (1 + x^2) + x|]$ + CONS\tan T