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

The integral of tan^2(x+c)+1 is tan(c+x)+C

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

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

tan (c + x) + C

Solution steps

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

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

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

Apply u-substitution

: \int sec^{2} (u) d u

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

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= tan (u)

Substitute back

u = c + x

= tan (c + x)

Add a constant to the solution

= tan (c + x) + C

x^{2}, 1

y^{^{''''}} + y^{^{''}} = 0

\frac{\partial}{\partial x} (x z - e^{x} sin (y))

x \to - 2 lim (\frac{5 x ^{2} + 6}{x})

(\frac{1}{( 1 - x ) ^{2}})^{^{'}}

What is the integral of tan^2(x+c)+1 ?
The integral of tan^2(x+c)+1 is tan(c+x)+C