How do you find the Maclaurin series of $f (x) = ln (1 + x^{2})$ $f(x)=ln(1+x^2)$ ?

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Answer 1 · 2015-11-05T02:02:01

Plug in $x^{2}$ $x^2$ for every x in the Maclaurin series for $f (x) = ln (1 + x)$ $f(x)=ln(1+x)$

So, you have the Maclaurin series for $f (x) = ln (1 + x)$ $f(x)=ln(1+x)$ defined as:

$f(x)=ln(1+x) = x-x^2/2+x^3/3-... =sum_(k=1)^n(((-1)^(k+1)x^k) / k)$

The question is essentially asking you to plug in $x^2$ for every x here, so:

$(x^2)^1-(x^2)^2/2+(x^2)^3/3-(x^2)^4/4+...$
which gives... $x^2 - x^4/2+x^6/3-x^8/4+... = sum_(k=1)^n(((-1)^(k+1)(x^2)^k) / k)$

How do you find the Maclaurin series of f(x)=ln(1+x^2)f(x)=ln(1+x2)f(x)=ln(1+x^2) ?