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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.

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  • $\begingroup$ You are on the right track. Could you establish the power series of $\frac{1}{4+x^2}$ ? $\endgroup$
    – Peter
    Oct 18 '15 at 19:43
  • $\begingroup$ If not, consider $\frac{1}{4+x^2}=\frac{1}{4}\frac{1}{1+\frac{x^2}{4}}=\frac{1}{4} \frac{1}{1-(-\frac{x^2}{4})}$ to get a geometric series again. $\endgroup$
    – Peter
    Oct 18 '15 at 19:44
  • $\begingroup$ Naive question: why do you have to use power series? I.e., for instance, why not use integration by parts, or another technique? $\endgroup$
    – Clement C.
    Oct 18 '15 at 19:49
  • $\begingroup$ Yes I believe I did get the series which I found to be $$\sum_{n=0}^{\infty} (-1)^n \frac{1}{4} (\frac{x^2}{4})^n$$ I am pretty sure that is correct, and also Clement C. I have to use it because that is what the section in the book is asking for. $\endgroup$
    – Ramirez77
    Oct 18 '15 at 19:51
  • $\begingroup$ OK. You have a factor $\frac{1}{n}$ missing, and the series should probably start at $n=1$... do you know the power series for $\ln(1+x)$? $\endgroup$
    – Clement C.
    Oct 18 '15 at 19:53
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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 $0 \leq y< 1$): $$ \ln(4+x^2) = 2\ln 2 - \sum_{n=1}^\infty \frac{(-1)^n}{n2^{2n}}x^{2n}. $$

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Why not to use just integration by parts?

$$ \int_{\frac{1}{2}}^{2}\log(4+x^2)\,dx = \left. x\log(4+x^2)\right|_{\frac{1}{2}}^{2} - \int_{\frac{1}{2}}^{2}\frac{2x^2}{4+x^2}\,dx $$ gives: $$ \int_{\frac{1}{2}}^{2}\log(4+x^2)\,dx = 6\log 2 +\frac{1}{2}\log\left(\frac{17}{4}\right)-4\int_{\frac{1}{4}}^{1}\frac{u^2}{1+u^2}\,du $$ hence: $$ \int_{\frac{1}{2}}^{2}\log(4+x^2)\,dx = 5\log(2)+\log\sqrt{17}-3+4\left(\frac{\pi}{4}-\arctan\frac{1}{4}\right). $$

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  • $\begingroup$ Apparently, the OP does not want to use IPP (the exercise being in the "Power series" section of a book). (cf. comments below the original question.) $\endgroup$
    – Clement C.
    Oct 18 '15 at 20:01
  • $\begingroup$ could younot use Power series of $5\log(2)+\log\sqrt{17}-3+4(\frac{\pi}{4}-\arctan\frac{1}{4})$? first three sums are done that leaves power series of$4(\frac{\pi}{4}-\arctan\frac{1}{4})$ $\endgroup$
    – onepound
    Dec 19 '18 at 15:02

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