Sabendo que integrate 1/(cos u) du = ln(|tan u + sec(u)|) + C determine, após fazer a substituição x = tan u a expressão da integral (sqrt(x ^ 2 + 1))/(pi ^ 2) A expressão procurada é ln(|x + sqrt(x ^ 2 + 1)|) - (sqrt(x ^ 2 + 1) / x) + C 8 B ln| 1/(1 + x ^ 2) + sqrt(x ^ 2 + 1) ]-(x/ sqrt (x^ 2 +1) )+C ln| 1 x + epsilon sqrt |x^ 2 +1 | - (sqrt(x ^ 2 + 1) / x) + C D ln(|x ^ 2 + sqrt(x + 1)|) - (sqrt((x + 1) / (x ^ 2))) + C ln(|x + x ^ 2 + 1|) - 1/x + C
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[tex]\displaystyle \sf \int \sqrt{x^2+1}dx \ \ ;\ \ x= tg(u) \to dx = sec^2(u)du \\\\\ \int \sqrt{tg^2(u)+1}\left(sec^2(u)du\right) \to \int\sqrt{sec^2(u)}\left(sec^2(u)du\right) \\\\\\ \int sec(u)\cdot sec^2(u)du =\int sec^3(u)du\to \\\\ \int sec(u)\cdot sec^2(u)du \to \int p\ dv = p\cdot v-\int v\ dp \\\\\ \text{Fa\c camos}:\\\\ p=sec(u) \to dp = sec(u)tg(u)du \\\\ dv = sec^2(u)du \to v = tg(u) \\\\ Da{\'i}}: \\\\ \int sec^3(u)du = sec(u)tg(u)-\int tg(u)sec(u)tg(u)du[/tex]
[tex]\displaystyle \sf \int sec^3(u)du = sec(u)tg(u)-\int tg^2(u)sec(u)du \\\\ \int sec^3(u)du = sec(u)tg(u)-\int \left(sec^2(u)-1\right)sec(u)du \\\\ \int sec^3(u)du = sec(u)tg(u)-\int sec^3(u)du+\int sec(u)du \\\\ 2\int sec^3(u)du = sec(u)tg(u)+\int sec(u)du \\\\\\\ \text{sabemos que} : \\\\ \int sec(u)du = \ln|sec(u)+tg(u)| +C \\\\\ Assim: \\\\ 2\int sec^3(u)du = sec(u)tg(u)+\ln|sec(u)+tg(u)| +C \\\\ \int sec^3(u)du = \frac{1}{2}sec(u)tg(u)+\frac{1}{2}\ln|sec(u)+tg(u)| +C[/tex]
[tex]\displaystyle \sf \text{desfazendo a substitui\c c\~ao} : \\\\ x = tg(u) \to sec^2(u) = tg^2(u)+1 \to sec(u) = \sqrt{x^2+1} \\\\\ Portanto: \\\\ \large\boxed{\sf \ \int \sqrt{x^2+1}\ dx = \frac{1}{2}\sqrt{x^2+1}\cdot x+\frac{1}{2}\ln\left|\sqrt{x^2+1}+x\right| +C \ }\checkmark[/tex]