Question

calcule a integral ∫√(sec⁴x+1)dx


obs: sec⁴x+1 estão dentro da raiz

Answer 1

$\displaystyle \int\sqrt{\sec^4(x+1)}dx=\int(\sec^4(x+1))^{\frac{1}{2}}dx=\int\sec^{4.\frac{1}{2}}(x+1)dx\implies\\\\ \int\sec^2(x+1)dx=\int\sec^2(u)du=\tan(u)=\tan(x+1)$

Answer 2

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$\displaystyle\sf\int\sqrt{sec^4(x+1)}dx=\int\diagdown\!\!\!\!\!\sqrt{sec^2(x+1))^{\diagup\!\!\!2}}=\int sec^2(x+1)dx\\\underline{\rm fac_{\!\!,}a}~t=x+1\implies dt=dx\\\displaystyle\sf\int sec^2(x+1)dx=\int sec^2(t)dt=tg(t)+k\\\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int\sqrt{sec^2(x+1)}dx=tg(x+1)+k}}}}\checkmark$