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

1 Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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