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

Nov 5, 2015

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

#### Explanation:

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

$f \left(x\right) = \ln \left(1 + x\right) = x - {x}^{2} / 2 + {x}^{3} / 3 - \ldots = {\sum}_{k = 1}^{n} \left(\frac{{\left(- 1\right)}^{k + 1} {x}^{k}}{k}\right)$

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

${\left({x}^{2}\right)}^{1} - {\left({x}^{2}\right)}^{2} / 2 + {\left({x}^{2}\right)}^{3} / 3 - {\left({x}^{2}\right)}^{4} / 4 + \ldots$
which gives... ${x}^{2} - {x}^{4} / 2 + {x}^{6} / 3 - {x}^{8} / 4 + \ldots = {\sum}_{k = 1}^{n} \left(\frac{{\left(- 1\right)}^{k + 1} {\left({x}^{2}\right)}^{k}}{k}\right)$