If derivative f'(x)=x/(2x^2+3x+1), what is the function f(x)?

To determine the original function, we'll have to
determine the indefinite integral of f'(x).

But first, we
must decompose the expression of f'(x) into partial
fractions:

We'll write the denominator as a product of
linear factors:

2x^2 + 3x + 1 =
(x+1)(2x+1)

x/(x+1)(2x+1) = A/(x+1) +
B/(2x+1)

x = A(2x+1) +
B(x+1)

We'll remove the
brackets:

x = 2Ax + A + Bx +
B

x = x(2A+B) + A +
B

Comparing, we'll
get:

2A+B=1

A + B=0 =>
A=-B

-2B+B=1

-B=1

B=-1
=> A=1

x/(x+1)(2x+1) = 1/(x+1) -
1/(2x+1)

Int[xdx/(x+1)(2x+1)]=Intdx/(x+1)-Int[dx/(2x+1)]

Intdx/(x+1)
= ln(x+1)

-Int[dx/(2x+1)] =
-[ln(2x+1)]/2

Int [xdx/(x+1)(2x+1)] = ln(x+1)-[ln(2x+1)]/2
+ C

The requested primitive function is: F(x)
= ln(x+1)-[ln(2x+1)]/2 + C