## Thomas' Calculus 13th Edition

$\Sigma_{n=1}^\infty \dfrac{ (-1)^{n-1}x^{(2n+1)}}{3n}$
Here, $\dfrac{x^3}{3} \cdot \ln (1+x^2)=\dfrac{x^3}{3}[x^2-\dfrac{(x^2)^2}{2!}+\dfrac{(x^2)^3}{3!}-...]$ This implies that $\dfrac{x^3}{3}[x^2-\dfrac{(x^2)^2}{2!}+\dfrac{(x^2)^3}{3!}-...]=\dfrac{x^3}{3}-\dfrac{x^5}{6}+\dfrac{x^7}{9}-...$ Hence, $\dfrac{x^3}{3} \cdot \ln (1+x^2)=\Sigma_{n=1}^\infty \dfrac{ (-1)^{n-1}x^{(2n+1)}}{3n}$