Integration problem:
If I Integrate this $$\ \int\frac{y}{(y+1)^2}dy $$ using integration by parts I get:$$\frac{-y}{y+1}+ln(y+1) $$

but if I do this: $$\ \int\frac{y}{(y+1)^2}dy $$
$$\ \int\frac{y+1}{(y+1)^2} -\frac{1}{(y+1)^2}dy $$and integrate I get
$$\ \frac{1}{y+1} +ln(y+1)$$

Sooo, what did I do wrong, what am I missing?

Question

Integration problem:
If I Integrate this  $$\ \int\frac{y}{(y+1)^2}dy $$ using integration by parts I get:$$\frac{-y}{y+1}+ln(y+1) $$

but if I do this: $$\ \int\frac{y}{(y+1)^2}dy $$
$$\ \int\frac{y+1}{(y+1)^2} -\frac{1}{(y+1)^2}dy $$and integrate I get 
$$\ \frac{1}{y+1} +ln(y+1)$$

Sooo, what did I do wrong, what am I missing?

anonymous · Answer

Dude can't you see both are one and the same thing

anonymous · Answer

Write -y as -y+1-1 see

anonymous · Answer

the second one is correct

lalaly · Answer

id prefer to use partial fractions or the second method u used

anonymous · Answer

no need for integration by parts here

anonymous · Answer

Just add constant of integration to both you would be surprised that after carring out what i just said you find same function with different constant of integration....
A result is consistent with what you expect out of an indefinite integration

anonymous · Answer

Method never changes the result it just changes the convinence of solving...

anonymous · Answer

Ah yes, thank you very much, this is actually a part of a bigger problem (a differential equation with given initial conditions) I didn't get the same constant C for the particular solution so i got confused, but it all works out nicely :D.

anonymous · Answer

"I didn't get the same constant C" like they got in the book.

anonymous · Answer

you said it's  part of a bigger problem...my instructor spent 30 min today stressing on the importance of adding the C right away cuz waiting til the end will change it completely. for ex: dy/dx=y...add C right away and you get y=Ce^x instead of y=e^x+C