Exercícios Resolvidos de Substituição Simples

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Exercício 811234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677787980828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263264265266267268269270271272273274275276277278279280281282283284285286287288289290291292293294295296297298299300301302303304305306307308309310311312313314315316317318319320321322323324325326327328329330331332333334335336337338339340341342343344345346347348349350351352353354355356357358359360361362363364365366367368369370371372373374375376377378379380381382383384385386387388389390391392393394395396397398399400401402403404405406407408409410411412413414415416417418419420421422423424425426427428429430431432433434435436437438439440441442443444445446447448449450451452453454455456457458459460461462463464465466467468469470471472473474475476477478479480481482483484485486487488489490491492493494495496497498499500501502503504505506507508509510511512513514515516517518519520521522523524525526527528529530531532533534535536537538539540541542543544545546547548549550551552553554555556557558559560561562563564565566567568569570571572573574575576577578579580581582583584585586587588589590591592593594595596597598599600601602603604605

Enunciado

Calcule as integrais dos exercícios 1 a 24 .

15 .   ∫ 3 x e x d x

Passo 1

Normalmente, o mais recomendável é chamar u do cara que é mais esquisito. No nosso caso, os dois caras são esquisitinhos, então vamos tentar olhar pros dois termos e ver se isso dá certo:

u = 3 e x → d u = 3 e x ln ⁡ 3 e d x = 3 e x ln ⁡ 3 + 1 d x

Isso quer dizer que:

( 3 e ) x d x = 1 ln ⁡ 3 + 1 d u

Então

  ∫ 3 x e x d x →     ∫ 1 ln ⁡ 3 + 1 d u

∫ 1 ln ⁡ 3 + 1 d u = u ln ⁡ 3 + 1 + C

Passo 2

Lembrando que a nossa variável NÃO é u , portanto, precisamos voltar pro que tínhamos antes:

u = 3 e x

u ln ⁡ 3 + 1 + C = 3 e x ln ⁡ 3 + 1 + C

Resposta

1 5 .   3 e x ln ⁡ 3 + 1 + C

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