Exercícios Resolvidos de Substituição Simples

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Exercício 102123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263264265266267268269270271272273274275276277278279280281282283284285286287288289290291292293294295296297298299300301302303304305306307308309310311312313314315316317318319320321322323324325326327328329330331332333334335336337338339340341342343344345346347348349350351352353354355356357358359360361362363364365366367368369370371372373374375376377378379380381382383384385386387388389390391392393394395396397398399400401402403404405406407408409410411412413414415416417418419420421422423424425426427428429430431432433434435436437438439440441442443444445446447448449450451452453454455456457458459460461462463464465466467468469470471472473474475476477478479480481482483484485486487488489490491492493494495496497498499500501502503504505506507508509510511512513514515516517518519520521522523524525526527528529530531532533534535536537538539540541542543544545546547548549550551552553554555556557558559560561562563564565566567568569570571572573574575576577578579580581582583584585586587588589590591592593594595596597598599600601602603604605

Enunciado

Calcule as integrais:

a   ∫ 1 1 + e x d x

Passo 1

Substituímos esse denominador aí:

u = 1 + e x       d u = e x d x

Só que na função eu não tenho exatamente esse d u . Então, vou forçar o d x a ser escrito como u :

d x = d u e x = d u u - 1

Assim, teremos na função:

∫ 1 1 + e x d x = ∫ 1 u d u u - 1

Passo 2

Resolvemos por frações parciais:

A u + B u - 1 = A u - A + B u u u - 1 = 1 u u - 1

A = - 1       ;       B = 1

∫ 1 u u - 1 d u = ∫ - 1 u + 1 u - 1 d u

= - ln ⁡ u + ln ⁡ u - 1 = ln ⁡ u + 1 u

= ln ⁡ e x 1 + e x + C

Resposta

∫ 1 1 + e x d x = ln ⁡ e x 1 + e x + C

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