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f(x) - Campus Rio Pomba

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Então,<br />

1 x dx<br />

-I o x2 ± 1 —<br />

(v) 12 x e -X.2 +1 dx .<br />

1<br />

2 du/21 2 du1 — ln I u I<br />

2 1 u 2<br />

= — 1 (ln 2 — ln 1) = — 1 ln 2 .<br />

2<br />

Calculamos primeiro a integral indefinida 1= x e<br />

Introdução à integração 375<br />

2<br />

1<br />

2<br />

x + 1<br />

Fazendo u = —x2 + 1, temos du = —2x dx ou x dx = -- du • Assim,<br />

2<br />

= S eu — du — 1 s — 1<br />

— eu du<br />

2<br />

2<br />

Logo,<br />

2 —<br />

x e-x + 1 dx — 1<br />

.1<br />

2<br />

6.10 EXERCÍCIOS<br />

x2 + 1<br />

2<br />

eu + c<br />

1 c_x2 + + c .<br />

2<br />

—14 + 1 1 - 1 + 1 — 1 e 3 +<br />

C + —<br />

2 2 2 2<br />

2 2 2<br />

1. Calculando as integrais / 1 = f x2 dx , 12 = x dx e 13 = dx ,<br />

obtemos // = 7/3, /2 = 3/2 e I3 = 1. Usando estes resultados, encontrar o valor de:

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