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

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### 1 Answer

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**