# Math Help - separable equation

1. ## separable equation

$(x+1)\frac{dy}{dx}=x(y^2+1)$

i basically get how to do the separable equations, i change it to:

$\int{\frac{dy}{(y^2+1)}}=\int{\frac{xdx}{(x+1)}}$

and then i integrate both sides, i'm having trouble integrating the right side, i tried using integration by parts but it doesn't come out how the book has it. i think maybe there's an easier way but i can't remember. can someone help me with this?

2. Let u = x+1

then the integrand becomes (u-1)/u = 1 - 1/u

Or use long division to obtain 1 - 1/(x+1) directly

at any rate you get x - ln(x+1) + c aftger integrating

3. Originally Posted by Calculus26
Let u = x+1

then the integrand becomes (u-1)/u = 1 - 1/u

Or use long division to obtain 1 - 1/(x+1) directly

at any rate you get x - ln(x+1) + c aftger integrating
No need for long division!

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