To find the antiderivative of tan^2x +tan^4x.

Let y = tan^2x (1+tan^2x)

Then y = tan^2x (sec^2x).

Now let tan x = t.

Then d/x(tanx) = secx^2x.

So, dt = sec^2x dx

So Antiderevative tan^2x sec^2x dx = Int tan^2x (1+tan^2x) dx = t^2 dt

Int tan^2x(1+tan^2xdx) dx = t ^3/3 + constant, as Int x^n dx = (1/n+1) x^(n+1) + const.

Therefore Int tan^2x(1+tan^x) dx = (1/3)(tanx)^3 +C.

Therefore Int (tan^2x+tan^4x) dx = (1/3)(tanx)^3 +C.

To determine the antiderivative of the given function , we'll have to calculate the indefinite integral of the given function:

Int [(tan x)^4 + (tan x)^2]dx

We notice that we can factorize by (tan x)^2 and the integrand will become:

Int (tan x)^2*[(tan x)^2 + 1]dx

We'll solve the integral using substitution technique. We'll note tan x = t:

We'll differentiate both sides:

dx/(cos x)^2 = dt

We'll use the fundamental formula of trigonometry and we'll get:

(sin x)^2 + (cos x)^2 =1

We'll divide the relation by (cos x)^2:

(tan x)^2 + 1 = 1/(cos x)^2

We'll re-write the integral, substituting [(tan x)^2 + 1] by 1/(cos x)^2:

Int (tan x)^2*[(tan x)^2 + 1]dx = Int (tan x)^2*dx/(cos x)^2

Now, we'll re-write the integral replacing the variable x by t:

Int (tan x)^2*dx/(cos x)^2 = Int t^2*dt

Int t^2*dt = t^3/3 + C

We'll substitute t by tan x and we'll get the antiderivative of the function:

**Int [(tan x)^4 + (tan x)^2]dx = (tan x)^3/3 + C**