What is the solution for (dy)/(dx)= 1/(x^2-1) ?

The solution for (dy)/(dx)= 1/(x^2-1) is y=-1/2 ln(x+1)+1/2 ln(x-1)+c_{1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dx)= 1/(x^2-1)

\frac{d y}{d x} = \frac{1}{x ^{2} - 1}

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

Solution steps

\frac{d y}{d x} = \frac{1}{x ^{2} - 1}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = \frac{1}{x ^{2} - 1}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{1}{x ^{2} - 1} d x

\int \frac{1}{x ^{2} - 1} d x = - \frac{1}{2} ln (x + 1) + \frac{1}{2} ln (x - 1) + c_{1}

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

\int_{0}^{t} x x - t d x

x \to 0 lim (\frac{1}{f ( x )})

\frac{d}{d x} (k e^{k x})

x \to 0 lim (\frac{6 cos ( x ) - 6}{x})

x + 3, \frac{( x + 3 )}{2}

What is the solution for (dy)/(dx)= 1/(x^2-1) ?
The solution for (dy)/(dx)= 1/(x^2-1) is y=-1/2 ln(x+1)+1/2 ln(x-1)+c_{1}