What is e^{-ln(1/(x^2)+1)} ?

The solution to e^{-ln(1/(x^2)+1)} is (x^2)/(x^2+1)

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simplify e^{-ln(1/(x^2)+1)}

simplify

e^{- l n (\frac{1}{x ^{2}} + 1)}

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

Solution steps

e^{- l n (\frac{1}{x ^{2}} + 1)}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{l n (\frac{1}{x ^{2}} + 1)}}

e^{l n (\frac{1}{x ^{2}} + 1)} = \frac{1 + x ^{2}}{x ^{2}}

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

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

c m^{2} (2 x + h)^{2} \cdot 6

(t \cdot e^{- 5 t}) \cdot f (t)

2 (2 (- 1)^{n - 2})

7

What is e^{-ln(1/(x^2)+1)} ?
The solution to e^{-ln(1/(x^2)+1)} is (x^2)/(x^2+1)