What is the integral of sqrt(tan^2(x)+1) ?

The integral of sqrt(tan^2(x)+1) is ln|tan(x)+sec(x)|+C

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integral of sqrt(tan^2(x)+1)

\int tan^{2} (x) + 1 d x

ln ∣ tan (x) + sec (x) ∣ + C

Solution steps

\int tan^{2} (x) + 1 d x

Rewrite using trig identities

= \int sec^{2} (x) d x

sec^{2} (x) = (sec (x)),

assuming

sec (x) \geq 0

= \int sec (x) d x

Use the common integral:

\int sec (x) d x = ln ∣ tan (x) + sec (x) ∣

= ln ∣ tan (x) + sec (x) ∣

Add a constant to the solution

= ln ∣ tan (x) + sec (x) ∣ + C

ln (1 + cos (x))

2 y y^{^{'}} = x (y^{2} + 1), y (2) = 0

f (x) = x^{4} cos (x)

\int e^{- 2 x} d x

\frac{d y}{d t} + y = t

What is the integral of sqrt(tan^2(x)+1) ?
The integral of sqrt(tan^2(x)+1) is ln|tan(x)+sec(x)|+C