What is expand-(2xy)/((x^2+1)(e^y+ln(x^2+1))) ?

The solution to expand-(2xy)/((x^2+1)(e^y+ln(x^2+1))) is -(2xy)/(e^yx^2+x^2ln(x^2+1)+e^y+ln(x^2+1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand-(2xy)/((x^2+1)(e^y+ln(x^2+1)))

expand

- \frac{2 x y}{( x ^{2} + 1 ) ( e ^{y} + ln ( x ^{2} + 1 ) )}

- \frac{2 x y}{e ^{y} x ^{2} + x ^{2} ln ( x ^{2} + 1 ) + e ^{y} + ln ( x ^{2} + 1 )}

Solution steps

- \frac{2 x y}{( x ^{2} + 1 ) ( e ^{y} + ln ( x ^{2} + 1 ) )}

Expand

(x^{2} + 1) (e^{y} + ln (x^{2} + 1)) : e^{y} x^{2} + x^{2} ln (x^{2} + 1) + e^{y} + ln (x^{2} + 1)

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

\frac{1}{4} (\frac{2 - x}{3})^{3}

(x - 2 a x + a)^{2}

(\frac{x}{3} - \frac{2 y}{5}) (\frac{x}{3} + \frac{2 y}{5})

\frac{x ^{2} - x - 1}{( x + 5 ) ( x ^{2} + 4 )}

4 (e^{- x} x - 2 e^{- x})

What is expand-(2xy)/((x^2+1)(e^y+ln(x^2+1))) ?
The solution to expand-(2xy)/((x^2+1)(e^y+ln(x^2+1))) is -(2xy)/(e^yx^2+x^2ln(x^2+1)+e^y+ln(x^2+1))