What is the integral from 0 to 1 of e^xtan(e^x) ?

The integral from 0 to 1 of e^xtan(e^x) is diverges

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integral from 0 to 1 of e^xtan(e^x)

\int_{0}^{1} e^{x} tan (e^{x}) d x

diverges

Solution steps

\int_{0}^{1} e^{x} tan (e^{x}) d x

Apply u-substitution

: \int_{1}^{e} tan (u) d u

= \int_{1}^{e} tan (u) d u

Take the denominator(s) of

tan (u)

and compare to zero:

u = \frac{π}{2}

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{1}^{\frac{π}{2}} tan (u) d u + \int_{\frac{π}{2}}^{e} tan (u) d u

\int_{1}^{\frac{π}{2}} tan (u) d u =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

x \to \infty + lim (3^{- x})

x \to 1 lim (\frac{in ( x )}{x - 1})

y = x^{2} + 10 x

\frac{d y}{d x} = \frac{x ^{2} - 2 y}{x}, y (2) = 2

5 (2^{x})

What is the integral from 0 to 1 of e^xtan(e^x) ?
The integral from 0 to 1 of e^xtan(e^x) is diverges