Hello!

Let's transform the expression under the integral sign:

`x/(2x^2-2c)-c/(x+1)=(x^2+x-2cx^2+2c^2)/((2x^2-2c)(x+1))=(x^2(1-2c)+x+2c^2)/((2x^2-2c)(x+1))=`

`=(x^2(1-2c))/((2x^2-2c)(x+1))+(x+2c^2)/((2x^2-2c)(x+1)).`

The second fraction behaves as `1/x^2` at infinity and therefore the integral of it converges at infinity, even absolutely. Consider the first fraction, it must converge also for the sum to converge.

For `1-2c!=0,` i.e. for `c!=1/2,` this...

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Hello!

Let's transform the expression under the integral sign:

`x/(2x^2-2c)-c/(x+1)=(x^2+x-2cx^2+2c^2)/((2x^2-2c)(x+1))=(x^2(1-2c)+x+2c^2)/((2x^2-2c)(x+1))=`

`=(x^2(1-2c))/((2x^2-2c)(x+1))+(x+2c^2)/((2x^2-2c)(x+1)).`

The second fraction behaves as `1/x^2` at infinity and therefore the integral of it converges at infinity, even absolutely. Consider the first fraction, it must converge also for the sum to converge.

For `1-2c!=0,` i.e. for `c!=1/2,` this fraction behaves like `1/x` at infinity. More precisely, for `1-2cgt0,` i.e. `clt1/2,`

`(x^2(1-2c))/((2x^2-c)(x+1))gt=(1-2c)*(x^2)/(3x^2*2x)=(1-2c)/6*1/x` for sufficiently large `x.` So this integral diverges at infinity. The similar estimate works for `cgt1/2.`

So the only candidate is `c=1/2,` for this `c` only the integral converges at infinity. But we have to check also the other possible critical points at `[1, oo).` They are the points where the denominator equals zero.

One of them is where `x+1=0,` i.e. `x_1=-1.` It is outside `[1,oo).` The next is where `2x^2-2c=2x^2-1=0.` They are `x_2=1/sqrt(2)` and `x_3=-1/sqrt(2),` also outside the realm of integration. Good, no more critical points!

The answer: the only such `c` is **1/2**.