Integral de 1/(e^x + e^-x)

Resolución integral número 5.

[@wasimvillidad3000]

J = ∫e ͯ/[(e ͯ)² + 1]dx Let u = e ͯ, du = e ͯdx J = ∫1/(u² + 1)du = arctan(u) + c = arctan(e ͯ) + c which is equivalent to your answer since arctan(1/e ͯ) + arctan(e ͯ) = π/2.

[@jitendrathakran9613]

Your methods are fantastic, God bless

[@nikolascsgo]

just multiply both sides by e^x and it's a perfect textbook substitution integral leading to arctg(e^x)

[@anonimo_19079]

Muy buen método, aunque yo habría intentado hacer un cambio de variable antes

[@Manish_babu-ft4up]

This is wrong # NCERT BOOK

[@jesusdavidescobardizz4027]

Crack 💪🏽

[@theelectro15]

como haces los videos, con que software?

[@doge_the_cat]

use SE ONE rule on 1/i

[@georgestoica9744]

Sech(x)/2