# Need a find int_(.5)^2 ln(4+x^2)dx

Need a find ${\int }_{.5}^{2}ln\left(4+{x}^{2}\right)dx$ I understand that to find this value I need to find what the power series of $ln\left(4+{x}^{2}\right)$ and I know how to start this off since $\mathrm{ln}\left(1-x\right)=\int \frac{1}{1-x}$ which is $\frac{1}{1-x}$ is the geometric series.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

zupa1z
Starting with $\mathrm{ln}\left(1+x\right)=-\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{n}}{n}$ (for |x|<1), observe that
$\mathrm{ln}\left(4+{x}^{2}\right)=\mathrm{ln}4+\mathrm{ln}\left(1+\frac{{x}^{2}}{4}\right)=2\mathrm{ln}2+\mathrm{ln}\left(1+\frac{{x}^{2}}{4}\right).$
In particular, we can apply the result above to $\mathrm{ln}\left(1+y\right)$, with $y=\frac{{x}^{2}}{4}$ (note that $0\le x<2$ implies $0\le y<1$):
$\mathrm{ln}\left(4+{x}^{2}\right)=2\mathrm{ln}2-\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{n{2}^{2n}}{x}^{2n}.$