Question

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}`

Answer 1

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.