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


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  • $\begingroup$ What have you tried? $\endgroup$ – 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 $$ \ln (3x^2+4x+1)=\ln (1+x)+\ln(1+3x)$$

All you have to do is to add the twe series.


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