Given function

$f(x)=\frac{x^2+2x+1}{x^2+1}=\frac{(x+1)^2}{x^2+1}.$

We need to check where this function is one-one or many-one.

Compute derivative

Let $u=(x+1)^2,\quad v=x^2+1.$ Then $u'=2(x+1),\quad v'=2x.$

Using quotient rule, $f'(x)=\frac{u'v-uv'}{v^2}$ $=\frac{2(x+1)(x^2+1)-2x(x+1)^2}{(x^2+1)^2}.$

Factor numerator: $f'(x)=\frac{2(x+1)\big[(x^2+1)-x(x+1)\big]}{(x^2+1)^2}$ $=\frac{2(x+1)(x^2+1-x^2-x)}{(x^2+1)^2}$ $=\frac{2(x+1)(1-x)}{(x^2+1)^2}$ $=\frac{2(1-x^2)}{(x^2+1)^2}.$

Study monotonicity

Since $(x^2+1)^2>0$ for all $x$, the sign of $f'(x)$ depends on $1-x^2$.

• If $|x|<1$, then $1-x^2>0$, so $f'(x)>0$.
• If $|x|>1$, then $1-x^2<0$, so $f'(x)<0$.
• At $x=\pm1$, $f'(x)=0$.

Therefore:

• $f$ is decreasing on $(-\infty,-1)$,
• $f$ is increasing on $(-1,1)$,
• $f$ is decreasing on $(1,\infty)$.

Check each option

Option A: $f(x)$ is many-one in $(-\infty,-1)$

On $(-\infty,-1)$, $f$ is strictly decreasing, hence one-one, not many-one.

So, A is false.

Option B: $f(x)$ is one-one in $(-\infty,\infty)$

Since the function first decreases, then increases, then decreases, it is not monotonic on all $\mathbb R$.

Also, for example, $f(-2)=\frac{(-2+1)^2}{(-2)^2+1}=\frac{1}{5},$ $f(0)=\frac{(0+1)^2}{0^2+1}=1,$ and we can also find repeated values. In particular, $f(-1)=0,\quad f(1)=2,$ so the function clearly does not behave injectively over all $\mathbb R$.

More directly, solve for equal outputs: $\frac{(x+1)^2}{x^2+1}=1 \implies (x+1)^2=x^2+1 \implies 2x=0 \implies x=0,$ but many other output values occur at multiple points because of change in monotonicity.

Hence B is false.

Option C: $f(x)$ is one-one in $[1,\infty)$ but not in $(-\infty,\infty)$

On $[1,\infty)$, we have $f'(x)\le 0$, and for $x>1$, $f'(x)<0$. So $f$ is strictly decreasing on $[1,\infty)$, hence one-one there.

It is not one-one on all $\mathbb R$. For instance, $f(-3)=\frac{(-2)^2}{10}=\frac25,$ $f\left(-\frac13\right)=\frac{(2/3)^2}{1/9+1}=\frac{4/9}{10/9}=\frac25.$ Two different inputs give the same output, so $f$ is not one-one on $\mathbb R$.

Therefore, C is true.

Option D: $f(x)$ is many-one in $(1,\infty)$

On $(1,\infty)$, $f$ is strictly decreasing, hence one-one.

So, D is false.

Final conclusion

The only correct option is $\boxed{\text{C}}.$

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

Hence, they agree.