# Maclaurin series expansion of $f(x) = \ln(3x^2 +4x +1)$ [closed]

Can someone please explain how I do the following Maclaurin series? $$f(x) = \ln(3x^2 +4x +1)$$

## closed as off-topic by Nosrati, Ak19, Robert Shore, Hayk, cmkJun 17 at 20:17

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• What have you tried? – Luke Collins Jun 17 at 10:59

Hint: $$\ln(3x^2 + 4x + 1) = \ln(3x+1) + \ln(x+1)$$
Note that $$\ln (1+x)=\int \frac {dx}{1+x}=$$
$$\int (1-x+x^2-x^3+...)=x-x^2/2+x^3/3-x^4/4+...$$
Similarly $$\ln (1+3x)=3x-(3x)^2/2+(3x)^3/3-....$$
Now we get $$\ln (3x^2+4x+1)=\ln (1+x)+\ln(1+3x)$$