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,

$\int _{}^{}1+tan^{2}xdx=\int _{}^{}1+{\frac {sin^{2}x}{cos^{2}x}}dx=\int _{}^{}{\frac {cos^{2}x+sin^{2}x}{cos^{2}x}}dx\cos ^{2}x+sin^{2}x=1\int _{}^{}{\frac {1}{cos^{2}x}}dx=\int _{}^{}\sec ^{2}xdx=tanx+C$