# Integral of [(1+2y^2)/(3-y)]dy (obtained from a differential equation)

This question actually arises from this Differential Equations question: Find the family of solutions for:

$\displaystyle(1+2y^2)\frac{dy}{dx} + (3-y)\cos x = 0$

I ruled out the methods I've so far learned in class (linear, exact, homogeneous, Bernoulli) and decided that it was a separable equation (correct me if I'm wrong), to reach:

$\displaystyle\int\frac{1+2y^2}{3-y}dy = -\int\cos x dx$

Although I like to think of Calculus as a strength of mine, I'm having difficulty with this seemingly simple integral on the left side. I've tried multiple methods of integration by parts, into double/triple integration by parts... What am I missing? Thanks

Divide $2y^2+1$ by $-y+3$. We get $-2y-6+\frac{19}{-y+3}$. Now the integration should be straightforward.
• You are welcome. If we are integrating $P(y)/q(y)$, where $P$ and $Q$ are polynomials such that the degree of $P$ is greater than or equal to the degree of $Q$, the usual first step is division. Then one may need to use partial fractions, though here we do not. – André Nicolas Sep 20 '14 at 1:42
• Note that (sadly) after doing the calculation, we will get an implicit equation for $y$. But we will not be able to solve explicitly for $y$ in terms of $x$ using elementary functions. – André Nicolas Sep 20 '14 at 1:45