Answer
$${P_4}\left( x \right) = 2x - 2{x^2} + \frac{8}{3}{x^3} - 4{x^4}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln \left( {1 + 2x} \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 + 2x} \right)} \right] = \frac{2}{{1 + 2x}} = 2{\left( {1 + 2x} \right)^{ - 1}} \cr
& {f^{\left( 2 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {2{{\left( {1 + 2x} \right)}^{ - 1}}} \right] = 2\left( { - 1} \right){\left( {1 + 2x} \right)^{ - 2}}\left( 2 \right) = - 4{\left( {1 + 2x} \right)^{ - 2}} \cr
& {f^{\left( 3 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ { - 4{{\left( {1 + 2x} \right)}^{ - 2}}} \right] = - 4\left( { - 2} \right){\left( {1 + 2x} \right)^{ - 3}}\left( 2 \right) = 16{\left( {1 + 2x} \right)^{ - 3}} \cr
& {f^{\left( 4 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {16{{\left( {1 + 2x} \right)}^{ - 3}}} \right] = 16\left( { - 3} \right){\left( {1 + 2x} \right)^{ - 4}}\left( 2 \right) = - 96{\left( {1 + 2x} \right)^{ - 5}} \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)} \right) = 0 \cr
& {f^{\left( 1 \right)}}\left( 0 \right) = 2{\left( {1 + 2\left( 0 \right)} \right)^{ - 1}} = 2 \cr
& {f^{\left( 2 \right)}}\left( 0 \right) = - 4{\left( {1 + 2\left( 0 \right)} \right)^{ - 2}} = - 4 \cr
& {f^{\left( 3 \right)}}\left( 0 \right) = 16{\left( {1 + 2\left( 0 \right)} \right)^{ - 3}} = 16 \cr
& {f^{\left( 4 \right)}}\left( 0 \right) = - 96{\left( {1 + 2\left( 0 \right)} \right)^{ - 5}} = - 96 \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{2}{{1!}}x + \frac{{ - 4}}{{2!}}{x^2} + \frac{{16}}{{3!}}{x^3} + \frac{{ - 96}}{{4!}}{x^4} \cr
& {\text{simplify }} \cr
& {P_4}\left( x \right) = 2x - 2{x^2} + \frac{8}{3}{x^3} - 4{x^4} \cr} $$
