how do you decompose 3x^2+7x+1/x(x+1)^2 into partial fractions?

Question

anonymous · Answer

$$\frac{3x^2+7x+1}{x(x+1)^2}$$
$$\frac{a}{x}+\frac{b}{x+1}+\frac{c}{(x+1)^2}$$ is a start

anonymous · Answer

then maybe 
$$a(x+1)^2+bx(x+1)+cx=3x^2+7x+1$$ you get $a$ immediately by replacing $x$ by 0

anonymous · Answer

in fact you get $a=1$ since the left side is $a$ and the right side is $1$

anonymous · Answer

now replace $x$ by $-1$ and get 
$$-c=-3$$ so $c=3$

anonymous · Answer

I can't wait to learn this properly.

anonymous · Answer

@eliassaab  has a snap way to do this that i cannot remember, but now that we have $a=1$ and $c=3$ we can find $b$ any number of ways. we see that the first term gives $x^2$ and the second will give $bx^2$ which together must be $3x^2$ making $b=2$

anonymous · Answer

i am not sure what "properly" means but there are many ways, on involving derivatives and the other limits, but i forget them. this is the easiest way i know how
you can also multiply out and equate like coefficients, but usually that is a long drag
perhaps not in this case

anonymous · Answer

i dont understand how you went from the first part to a(x+1)^2+bx(x+1)+cx=3x^2+7x+1

anonymous · Answer

add the fractions and that will be your numerator

anonymous · Answer

@satellite73 : I mean't I can't wait till I learn this in Class. So far I learnt Partial fractions through online videos.