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

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

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expand e^{2(x+1-ln|x+1|)}

expand

e^{2 (x + 1 - l n ∣ x + 1 ∣)}

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

Solution steps

e^{2 (x + 1 - l n ∣ x + 1 ∣)}

Expand

2 (x + 1 - ln ∣ x + 1 ∣) : 2 x + 2 - 2 ln ∣ x + 1 ∣

= e^{2 x + 2 - 2 l n ∣ x + 1 ∣}

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

= e^{- 2 l n ∣ x + 1 ∣} e^{2 x + 2}

Simplify

e^{- 2 l n ∣ x + 1 ∣} : ∣ x + 1 ∣^{- 2}

= ∣ x + 1 ∣^{- 2} e^{2 x + 2}

Apply exponent rule:

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

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

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

Apply absolute rule:

∣ x ∣^{2} = x^{2}

, assuming

x \in R

∣ x + 1 ∣^{2} = (x + 1)^{2}

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

Simplify

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

(x^{4} - 2 x^{2} + 2) (x^{2} - 2 x + 3)^{'}

(2, 0, 2,) \cdot (x, 4, - 1) \cdot (3, 2, 1)

\frac{d ^{2}}{d x ^{2}} ((y) (2 x^{2} - 4))

(2 + 3 i) (2 - 3 i)^{'}

x (1 - x - x^{3} + x^{4})^{2}

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