## University Calculus: Early Transcendentals (3rd Edition)

$x^2-\dfrac{x^4}{2}+\dfrac{x^{6}}{3}-\dfrac{x^{8}}{4}$
Since, we know that the Maclaurin Series for $\ln(1+x)$ is defined as: $\ln (1+x)=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^{n+1}}{(n+1)!}=x-\dfrac{x^2}{2!}+\dfrac{x^3}{3!}-\dfrac{x^4}{4!}+...$ Now, $\ln (1+x^2)=x^2-\dfrac{(x^2)^2}{2!}+\dfrac{(x^2)^3}{3!}-\dfrac{(x^2)^4}{4!}+...$ or, $=x^2-\dfrac{x^4}{2}+\dfrac{x^{6}}{3}-\dfrac{x^{8}}{4}$