Answer
$${P_4}\left( x \right) = 2{x^2} - 2{x^4}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln \left( {1 + 2{x^2}} \right) \cr
& {\text{Use the definition of Taylor Polynomial of Degree }}n\,\,\,\left( {{\text{see page 629}}} \right) \cr
& {\text{Let }}f{\text{ be a function that can be differentiated }}n{\text{ times at 0}}{\text{. The Taylor }} \cr
& {\text{polynomial of degree }}n{\text{ for }}f{\text{ at 0 is }} \cr
& {P_n}\left( x \right) = f\left( 0 \right) + \frac{{{f^{\left( 1 \right)}}\left( 0 \right)}}{{1!}}x + \frac{{{f^{\left( 2 \right)}}\left( 0 \right)}}{{2!}}{x^2} + \cdots + \frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n} = \sum\limits_{i = 0}^n {\frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{i!}}} {x^i} \cr
& {\text{Find the Taylor polynomials of degree 4 at 0}}{\text{. }} \cr
& {\text{then }}n = 4. \cr
& {\text{The }}n{\text{ - th derivatives are}} \cr
& {f^{\left( 1 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {\ln \left( {1 + 2{x^2}} \right)} \right] = \frac{{4x}}{{1 + 2{x^2}}} \cr
& {f^{\left( 2 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{4x}}{{1 + 2{x^2}}}} \right] \cr
& {\text{by using the quotient rule}} \cr
& {f^{\left( 2 \right)}}\left( x \right) = \frac{{\left( {1 + 2{x^2}} \right)\frac{d}{{dx}}\left[ {4x} \right] - \left( {4x} \right)\frac{d}{{dx}}\left[ {1 + 2{x^2}} \right]}}{{{{\left( {1 + 2{x^2}} \right)}^2}}} = \frac{{\left( {1 + 2{x^2}} \right)\left( 4 \right) - 4x\left( {4x} \right)}}{{{{\left( {1 + 2{x^2}} \right)}^2}}} \cr
& {f^{\left( 2 \right)}}\left( x \right) = \frac{{4 + 8{x^2} - 16{x^2}}}{{{{\left( {1 + 2{x^2}} \right)}^2}}} = \frac{{4 - 8{x^2}}}{{{{\left( {1 + 2{x^2}} \right)}^2}}} \cr
& {f^{\left( 3 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{4 - 8{x^2}}}{{{{\left( {1 + 2{x^2}} \right)}^2}}}} \right] \cr
& {\text{by using the quotient rule}} \cr
& {f^{\left( 3 \right)}}\left( x \right) = \frac{{{{\left( {1 + 2{x^2}} \right)}^2}\frac{d}{{dx}}\left[ {4 - 8{x^2}} \right] - \left( {4 - 8{x^2}} \right)\frac{d}{{dx}}\left[ {{{\left( {1 + 2{x^2}} \right)}^2}} \right]}}{{{{\left( {1 + 2{x^2}} \right)}^4}}} \cr
& {f^{\left( 3 \right)}}\left( x \right) = \frac{{{{\left( {1 + 2{x^2}} \right)}^2}\left( { - 16x} \right) - 2\left( {4 - 8{x^2}} \right)\left( {1 + 2{x^2}} \right)\left( {4x} \right)}}{{{{\left( {1 + 2{x^2}} \right)}^4}}} \cr
& {f^{\left( 3 \right)}}\left( x \right) = \frac{{\left( {1 + 2{x^2}} \right)\left( { - 16x} \right) - 2\left( {4 - 8{x^2}} \right)\left( {4x} \right)}}{{{{\left( {1 + 2{x^2}} \right)}^3}}} = \frac{{ - 16x - 32{x^3} - 32x + 64{x^3}}}{{{{\left( {1 + 2{x^2}} \right)}^3}}} \cr
& {f^{\left( 3 \right)}}\left( x \right) = \frac{{32{x^3} - 48x}}{{{{\left( {1 + 2{x^2}} \right)}^3}}} \cr
& {f^{\left( 4 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{32{x^3} - 48x}}{{{{\left( {1 + 2{x^2}} \right)}^3}}}} \right] \cr
& {\text{by using the quotient rule}} \cr
& {f^{\left( 4 \right)}}\left( x \right) = \frac{{{{\left( {1 + 2{x^2}} \right)}^3}\left( {96{x^2} - 48} \right) - \left( {32{x^3} - 48x} \right)\left( 3 \right){{\left( {1 + 2{x^2}} \right)}^2}\left( {4x} \right)}}{{{{\left( {1 + 2{x^2}} \right)}^6}}} \cr
& {f^{\left( 4 \right)}}\left( x \right) = \frac{{\left( {1 + 2{x^2}} \right)\left( {96{x^2} - 48} \right) - \left( {32{x^3} - 48x} \right)\left( 3 \right)\left( {4x} \right)}}{{{{\left( {1 + 2{x^2}} \right)}^4}}} \cr
& {f^{\left( 4 \right)}}\left( x \right) = \frac{{96{x^2} - 48 + 192{x^4} - 96{x^2} - 384{x^4} + 576{x^2}}}{{{{\left( {1 + 2{x^2}} \right)}^4}}} \cr
& {f^{\left( 4 \right)}}\left( x \right) = \frac{{576{x^2} - 192{x^4} - 48}}{{{{\left( {1 + 2{x^2}} \right)}^4}}} \cr
& \cr
& {\text{evaluate }}f\left( 0 \right),{f^{\left( 1 \right)}}\left( 0 \right),{f^{\left( 2 \right)}}\left( 0 \right),{f^{\left( 3 \right)}}\left( 0 \right),{f^{\left( 4 \right)}}\left( 0 \right) \cr
& f\left( 0 \right) = \ln \left( {1 + 2{{\left( 0 \right)}^2}} \right) = 0 \cr
& {f^{\left( 1 \right)}}\left( 0 \right) = \frac{{4\left( 0 \right)}}{{1 + 2{{\left( 0 \right)}^2}}} = 0 \cr
& {f^{\left( 2 \right)}}\left( 0 \right) = \frac{{4 - 8{{\left( 0 \right)}^2}}}{{{{\left( {1 + 2{{\left( 0 \right)}^2}} \right)}^2}}} = 4 \cr
& {f^{\left( 3 \right)}}\left( 0 \right) = \frac{{32{{\left( 0 \right)}^3} - 48\left( 0 \right)}}{{{{\left( {1 + 2{{\left( 0 \right)}^2}} \right)}^3}}} = 0 \cr
& {f^{\left( 4 \right)}}\left( 0 \right) = \frac{{576{{\left( 0 \right)}^2} - 192{{\left( 0 \right)}^4} - 48}}{{{{\left( {1 + 2{{\left( 0 \right)}^2}} \right)}^4}}} = - 48 \cr
& {\text{Replace the found values into the definition of Taylor Polynomial of Degree }}n \cr
& {\text{for }}n = 4 \cr
& {P_4}\left( x \right) = f\left( 0 \right) + \frac{{{f^{\left( 1 \right)}}\left( 0 \right)}}{{1!}}x + \frac{{{f^{\left( 2 \right)}}\left( 0 \right)}}{{2!}}{x^2} + \frac{{{f^{\left( 3 \right)}}\left( 0 \right)}}{{3!}}{x^3} + \frac{{{f^{\left( 4 \right)}}\left( 0 \right)}}{{4!}}{x^4} \cr
& {P_4}\left( x \right) = 0 + \frac{0}{{1!}}x + \frac{4}{{2!}}{x^2} + \frac{0}{{3!}}{x^3} + \frac{{ - 48}}{{4!}}{x^4} \cr
& {\text{simplify }} \cr
& {P_4}\left( x \right) = 2{x^2} - 2{x^4} \cr} $$
