Question

if x+1/x= 2 then
$x { \: }^{2} + 1 \: \: \: \div \ \: x {}^{2}$
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Answer 1

Answer:

This is an old favorite of mine.

Define

I=∫+∞−∞e−x2dx

Then

I2=(∫+∞−∞e−x2dx)(∫+∞−∞e−y2dy)

I2=∫+∞−∞∫+∞−∞e−(x2+y2)dxdy

Now change to polar coordinates

I2=∫+2π0∫+∞0e−r2rdrdθ

The θ integral just gives 2π, while the r integral succumbs to the substitution u=r2

I2=2π∫+∞0e−udu/2=π

So

I=π−−√

and your integral is half this by symmetry

I have always wondered if somebody found it this way, or did it first using complex variables and noticed this would work.

Answer 2

Step-by-step explanation:

x+1/x= 2

squaring in both side

(x+1/x)²=2²

or (x-1/x)²+4 x×1/x=4

or (x-1/x)²+4=4

or, (x-1/x)²=0

or x-1/x=0

0r x=1/x

or x²=1

or x=1

than x²+1/x²

= 1²+1/1²

=1+1

=2. (ans)