Tangent integral

$∫(\tan x(\tan x + \sec x))dx$

$∫(\tan^{2} x + \tan x \sec x)dx$

$∫(\sec^{2} x -1 + \tan x \sec x)dx$

Since $∫\sec^{2} x dx=\tan x + c$ and $∫\tan x \sec x-1)dx=\sec x$ so...

$=\boxed{\tan x -x+ \sec x+C}$

Take the integral: integral tan(x) (tan(x)+sec(x)) dx

Expanding the integrand tan(x) (tan(x)+sec(x)) gives tan^2(x)+tan(x) sec(x): = integral (tan^2(x)+tan(x) sec(x)) dx

Integrate the sum term by term: = integral tan^2(x) dx+ integral tan(x) sec(x) dx

Rewrite tan(x) sec(x) as (sin(x))/(cos^2(x)): = integral tan^2(x) dx+ integral (sin(x))/(cos^2(x)) dx

For the integrand (sin(x))/(cos^2(x)), substitute u = cos(x) and du = -sin(x) dx: = integral -1/u^2 du+ integral tan^2(x) dx

Factor out constants: = integral tan^2(x) dx- integral 1/u^2 du

The integral of 1/u^2 is -1/u: = 1/u+ integral tan^2(x) dx

Write tan^2(x) as sec^2(x)-1: = 1/u+ integral (sec^2(x)-1) dx

Integrate the sum term by term and factor out constants: = 1/u- integral 1 dx+ integral sec^2(x) dx

The integral of sec^2(x) is tan(x): = 1/u+tan(x)- integral 1 dx

The integral of 1 is x: = 1/u-x+tan(x)+constant

Substitute back for u = cos(x): Answer: = -x+tan(x)+sec(x)+constant