How does $\frac{x^2}{ x^2 + 1}$ simplify to $1-\frac{1}{1+x^2}$

Question

How does $$\frac{x^2}{ x^2 + 1}\quad\text{ simplify to }\quad 1-\frac{1}{1+x^2}\;?$$

Can someone explain the steps of how to get to that alternate form?

Answer 1

$$\frac{x^2}{1+x^2}=\frac{1+x^2-1}{1+x^2}=\frac{1+x^2}{1+x^2}-\frac1{1+x^2}=1-\frac1{1+x^2}$$

Answer 2

$\frac{x^2}{x^2+1}+\frac{1}{x^2+1}=\frac{x^2+1}{x^2+1}=1$

So

$1-\frac{1}{x^2+1}=\frac{x^2}{x^2+1}$

For example if $x=2$ then

$\frac{4}{5}+\frac{1}{5}=\frac{5}{5}=1$

$1-\frac{1}{5}=\frac{4}{5}$

Answer 3

@André Nicolas already said this, but you need to use polynomial long division. When you divide $x^2$ by $x^2 + 1$, you get 1, and when you subtract $x^2 + 1$ from $x^2$, you get -1. Since that is your remainder, your final answer is: $$1 - \frac {1}{x^2 + 1}$$

Answer 4

If a rational function has a numerator of degree not strictly less than the denominator, you can apply polynomial Euclidean division of numerator (here $x^2$) by denominator ($x^2+1)$. The quotient ($1$) becomes a polynomial part, the remainder ($-1$) becomes the new numerator, over an unchanged denominator.

In formula, if $A,B$ are polynomials, $B\neq0$ and $A=QB+R$ by Euclidean division, then $$ \frac AB=Q+\frac RB\quad. $$