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Find all the values of x in the set of complex numbers that satisfy the following equation:


\boxed{\sum^{\lceil \int^{(\pi)/(4)}_02secxdx\rceil }_(k=\lfloor \int^2_0lnxdx\rfloor)((d)/(dx)(x^(k+2)))=-\lceil lim_(a\to\infty)\int_(-a)^a(1)/(x^2+1)dx\rceil!+1}

1 Answer

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Compute all the component integrals first:


I_1=\displaystyle\int_0^(\pi/4)2\sec x\,\mathrm dx=2\ln(\sqrt2+1)

I_2=\displaystyle\int_0^2\ln x\,\mathrm dx=2(\ln2-1)

I_3=\displaystyle\lim_(a\to\infty)\int_(-a)^a(\mathrm dx)/(x^2+1)=\pi

Now,


\sqrt2\approx1.4\implies \sqrt2+1\approx2.4<e\implies 2\ln(\sqrt2+1)<2\ln e=2

\implies \left\lceil I_1\right\rceil=2


e<4<e^2\implies1<\ln4<2\implies-1<2(\ln2-1)<0

\implies\left\lfloor I_2\right\rfloor=-1


\pi\approx3.14\implies\left\lceil I_3\right\rceil=4

So the given equation reduces to


\displaystyle\sum_(k=-1)^2(\mathrm d)/(\mathrm dx)x^(k+2)=1-4!

(\mathrm dx)/(\mathrm dx)+(\mathrm dx^2)/(\mathrm dx)+(\mathrm dx^3)/(\mathrm dx)+(\mathrm dx^4)/(\mathrm dx)=-23

4x^3+3x^2+2x+24=0

a fairly standard cubic. Incidentally, when
x=-2, the LHS reduces to 0, so
x+2 is a factor of the cubic. You can find the remaining two solutions easily with the quadratic formula.
User Nabeelmukhtar
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