I have a question that is asking for me to find $\int_{.5}^{2} ln(4+x^2) dx$ I understand that to find this value I need to find what the power series of $ln(4+x^2)$ and I know how to start this off since $\ln(1-x)= \int\frac{1}{1-x}$ which is $\frac{1}{1-x}$ is the geometric series.
My problem is that is that even with this knowledge I am stumped on how to continue through this problem. I am pretty sure finding out what the power series for $\frac{1}{4+x^2}$ should be and then integrating it should give me my series for $ln(4+x^2)$ but when I tried to do the integration my terms don't seem to match up with what is expected from them. Any help would be greatly appreciated.
Best Answer
Starting with $\ln(1+x) = -\sum_{n=1}^\infty (-1)^n \frac{x^n}{n}$ (for $\lvert x\rvert < 1$), observe that $$ \ln(4+x^2) = \ln 4 + \ln\left(1+\frac{x^2}{4}\right)= 2\ln 2 + \ln\left(1+\frac{x^2}{4}\right). $$ In particular, we can apply the result above to $\ln(1+y)$, with $y=\frac{x^2}{4}$ (note that $0 \leq x< 2$ implies $0 \leq y< 1$): $$ \ln(4+x^2) = 2\ln 2 - \sum_{n=1}^\infty \frac{(-1)^n}{n2^{2n}}x^{2n}. $$