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ln(1 + x^2) how would I do this? Thanks in advance.

Quote: Originally Posted by khuezy ln(1 + x^2) how would I do this? Thanks in advance. Remember $\displaystyle \ln (1 + t) = \sum_{n=0}^{\infty} \frac{(-1)^n t^n}{n}$ for $\displaystyle |t|<1$. Therefore if $\displaystyle x^2 < 1$ then $\displaystyle \ln (1+x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n}$