Expert: Socrates - 8/24/2006

Question
Hey I am doing an integral in diffy q. I was wondering if you could help me with the integral du/(u^2+2u)? The original problem was (y^2 +yx)dx + x^2dy = 0 and using y =ux and dy = udx + xdu it simplified to dx/x + du/(u^2 + 2u) = 0. I am just drawing a blank with what to do with the left integral.! thank you if you can help me

Answer
You want S 1/(u^2+2u) du

Use partial fractions

a/u + b/(u+2) = 1/(u)(u+2) = 1/(u^2+2u)

get a=1/2 , b=-1/2

so

S 1/(u^2+2u) du = S (1/2)/u  - (1/2)/(u+2) du =

(1/2) S 1/u - 1/(u+2) du =

(1/2) (lnu - ln(u+2)) =

(1/2) ln (u/(u+2))

good luck !