Find all the values of x in the set of complex numbers that satisfy the following equation:[tex]\boxed{\sum^{\lceil \int^{\frac{\pi}{4}}_02secxdx\rceil }_{k=\lfloor \int^2_0lnxdx\rfloor}(\frac{d}{dx}(x^{k+2}))=-\lceil lim_{a\to\infty}\int_{-a}^a\frac{1}{x^2+1}dx\rceil!+1}[/tex]

Compute all the component integrals first:

[tex]I_1=\displaystyle\int_0^{\pi/4}2\sec x\,\mathrm dx=2\ln(\sqrt2+1)[/tex]
[tex]I_2=\displaystyle\int_0^2\ln x\,\mathrm dx=2(\ln2-1)[/tex]
[tex]I_3=\displaystyle\lim_{a\to\infty}\int_{-a}^a\frac{\mathrm dx}{x^2+1}=\pi[/tex]

Now,

[tex]\sqrt2\approx1.4\implies \sqrt2+1\approx2.4<e\implies 2\ln(\sqrt2+1)<2\ln e=2[/tex]
[tex]\implies \left\lceil I_1\right\rceil=2[/tex]

[tex]e<4<e^2\implies1<\ln4<2\implies-1<2(\ln2-1)<0[/tex]
[tex]\implies\left\lfloor I_2\right\rfloor=-1[/tex]

[tex]\pi\approx3.14\implies\left\lceil I_3\right\rceil=4[/tex]

So the given equation reduces to

[tex]\displaystyle\sum_{k=-1}^2\frac{\mathrm d}{\mathrm dx}x^{k+2}=1-4![/tex]
[tex]\dfrac{\mathrm dx}{\mathrm dx}+\dfrac{\mathrm dx^2}{\mathrm dx}+\dfrac{\mathrm dx^3}{\mathrm dx}+\dfrac{\mathrm dx^4}{\mathrm dx}=-23[/tex]
[tex]4x^3+3x^2+2x+24=0[/tex]

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