What is the integral of e^xcot^2(e^x+1) ?

The integral of e^xcot^2(e^x+1) is -e^x-1-cot(e^x+1)+C

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integral of e^xcot^2(e^x+1)

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

- e^{x} - 1 - cot (e^{x} + 1) + C

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

\int 1 d u = u

\int csc^{2} (u) d u = - cot (u)

= - u - cot (u)

Substitute back

u = e^{x} + 1

= - (e^{x} + 1) - cot (e^{x} + 1)

Simplify

= - e^{x} - 1 - cot (e^{x} + 1)

Add a constant to the solution

= - e^{x} - 1 - cot (e^{x} + 1) + C

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

\int \frac{( ln ^{4} ( x ) )}{x} d x

\int \frac{ln ( x + 1 )}{x} d x

What is the integral of e^xcot^2(e^x+1) ?
The integral of e^xcot^2(e^x+1) is -e^x-1-cot(e^x+1)+C