What is the integral of e^xtan^2(e^x) ?

The integral of e^xtan^2(e^x) is -e^x+tan(e^x)+C

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integral of e^xtan^2(e^x)

\int e^{x} tan^{2} (e^{x}) d x

- e^{x} + tan (e^{x}) + C

Solution steps

\int e^{x} tan^{2} (e^{x}) d x

Rewrite using trig identities

= \int e^{x} (- 1 + sec^{2} (e^{x})) d x

Apply u-substitution

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

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

Rewrite using trig identities

= \int - 1 + sec^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 1 d u + \int sec^{2} (u) d u

\int 1 d u = u

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

= - u + tan (u)

Substitute back

u = e^{x}

= - e^{x} + tan (e^{x})

Add a constant to the solution

= - e^{x} + tan (e^{x}) + C

\int \frac{5}{x ( x - 5 ) ^{2}} d x

f (t) = sin (4 t)

x \to π lim (x sec (x))

\int_{- 2}^{- 1} 4 x^{3} + \frac{2}{x ^{3}} d x

\frac{d}{d x} (\frac{2 e ^{3 x} - 3}{x + 1})

What is the integral of e^xtan^2(e^x) ?
The integral of e^xtan^2(e^x) is -e^x+tan(e^x)+C