5.4 Indefinite Integrals and the Net Change Theorem/17

$\int (1+\tan ^{2}{\alpha })\,d\alpha =\int \sec ^{2}\alpha \,d\alpha =\tan \alpha +C$

Note: $1+\tan ^{2}{\alpha }=\sec ^{2}\alpha$

Or,

$\int (1+\tan ^{2}{\alpha })\,d\alpha =\int \left(1+{\frac {sin^{2}\alpha }{cos^{2}\alpha }}\right)d\alpha =\int \left({\frac {cos^{2}\alpha +sin^{2}\alpha }{cos^{2}\alpha }}\right)d\alpha \cos ^{2}x+sin^{2}x=1\int {\frac {1}{cos^{2}x}}dx=\int \sec ^{2}xdx=\tan {x}+C$