Find the taylor series about x=0 for the following functiong(x) = ln(x^(2) +1)

Find the taylor series about x=0 for the following functiong(x) = ln(x^(2) +1)

Find the taylor series about x=0 for the following functiong(x) = ln(x^(2) +1)

Find the taylor series about x=0 for the following functiong(x) = ln(x^(2) +1)
If we try to take derivative after derivative, things will get messy quickly. Instead, we can use the following trick:`g(x)= “ln” (x^2+1)“g'(x)= (2x)/(x^2+1)`We can find the power series for `g'(x)` using what we know about geometric series:`1+z+z^2+z^3+… = (1)/(1-z)` (provided that `|z|<1` )If we say `z=-x^2`, we have:`1-x^2+x^4-x^6+…=(1)/(1+x^2)`So:`g'(x)= (2x)/(x^2+1)“=2x(1-x^2+x^4-x^6+…)“=2x-2x^3+2x^5-2x^7+…`We can integrate term by term to get `g(x)“g(x) = C + x^2 – (2/4)x^4 + (2/6)x^6-(2/8)x^8+…“=C+x^2 – (x^4)/2 + (x^6)/3 – (x^8)/4 + …`To get C, we plug in `x=0`:`g(0) = C + 0^2 + …“g(0) = “ln” (0^2+1) = 0`So `C=0`And`g(x)=x^2 – (x^4)/2 + (x^6)/3 – (x^8)/4 + …`Or, in summation notation:`g(x) = sum_(n=1)^oo (x^(2n))/n`