Đặt $x = \tan t$. Khi đó $dx = d(\tan t) = \dfrac{1}{\cos^2t} dt$, $t = \arctan x$.
Đặt $a = \dfrac{\pi}{4}, b = \arctan(2)$. KHi đó
$I = \int_a^b \dfrac{\sqrt{1 + \tan^2t}}{\tan t.\cos^2t} dt$
$= \int_a^b \dfrac{dt}{\cos^3t \tan t}$
$= \int_a^b \dfrac{dt}{\cos^2t \sin t}$
$= \int_a^b \dfrac{\sin t dt}{\sin^2t \cos^2t}$
$= -\int_a^b \dfrac{d(\cos t)}{\cos^2t (1 - \cos^2t)}$
Đặt $u = \cos t$, $c = \cos a, d = \cos b$. KHi đó
$I = - \int_c^d \dfrac{du}{u^2(1-u^2)}$
$= - \int_c^d \left( \dfrac{1}{1 - u^2} + \dfrac{1}{u^2} \right)du$
$= \int_c^d \left( \dfrac{1}{u^2-1} - \dfrac{1}{u^2} \right) du$
$= \int_c^d \left[ \dfrac{1}{2} \left( \dfrac{1}{u+1} - \dfrac{1}{u-1} \right) - \dfrac{1}{u^2} \right]du$
$= \dfrac{1}{2} ( \ln|u+1| - \ln|u-1|) + \dfrac{1}{u}\big\vert_c^d$
Do đó
$I = \dfrac{1}{2} \ln(\cos t+1) - \dfrac{1}{2} \ln(1 - \cos t) + \dfrac{1}{\cos t}$
$= \dfrac{1}{2} \ln[\cos(\arctan x) + 1] - \dfrac{1}{2} \ln[1 - \cos(\arctan x)] + \dfrac{1}{\cos(\arctan x)}\big\vert_1^2$
$= \dfrac{1}{2} \ln[\cos(\arctan 2) + 1] - \dfrac{1}{2} \ln[1 - \cos(\arctan 2)] + \dfrac{1}{\cos(\arctan 2)} - \left\{ \dfrac{1}{2} \ln[\dfrac{1}{\sqrt{2}} + 1] - \dfrac{1}{2} \ln[1 -\dfrac{1}{\sqrt{2}}] + \sqrt{2} \right\}$
