Question
$\int \sec ^2 x \sqrt{\tan ^2 x+\tan x-7} d x$

Answer

Let $I =\int \sec ^2 x \sqrt{\tan ^2 x+\tan x-7} d x$
Put $\tan x=t$
$ \therefore \sec ^2 x d x = dt$
$\therefore I =\int \sqrt{ t ^2+ t -7} dt$
$=\int \sqrt{ t ^2+ t +\frac{1}{4}-\frac{1}{4}-7} dt$
$=\int \sqrt{\left( t +\frac{1}{2}\right)^2-\frac{29}{4}} dt$
$=\int \sqrt{\left( t +\frac{1}{2}\right)^2-\left(\frac{\sqrt{29}}{2}\right)^2} dt$
$=\frac{ t +\frac{1}{2}}{2} \sqrt{\left( t +\frac{1}{2}\right)^2-\left(\frac{\sqrt{29}}{2}\right)^2}$
$=-\frac{\left(\frac{\sqrt{29}}{2}\right)^2}{2} \log \left| t +\frac{1}{2}+\sqrt{ t ^2+ t -7}\right|+ c$
$=\frac{2 t +1}{4} \sqrt{ t ^2+ t -7}-\frac{29}{8} \log \left| t +\frac{1}{2}+\sqrt{ t ^2+ t -7}\right|+ c$
$\therefore I =$
$\frac{(2 tan x+1)}{4} \sqrt{\tan ^2 x+\tan x-7}-\frac{29}{8} \log \left|\tan x+\frac{1}{2}+\sqrt{\tan ^2 x+\tan x-7}\right|+ c $

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