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: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 1+\tan^2{\alpha} = \sec^2\alpha}

Or,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int_{}^{}1+tan^2xdx = \int_{}^{}1+\frac{sin^2x}{cos^2x}dx = \int_{}^{}\frac{cos^2x+sin^2x}{cos^2x}dx \cos^2x+sin^2x=1 \int_{}^{}\frac{1}{cos^2x}dx = \int_{}^{}\sec^2xdx = tanx+C }