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I know what to do when we have something like x cos(x), you just compute it for cosx and then multiply the terms by x at the end, but what do you do when you have something like ln(3x^2+1)?

Do you replace 3x^2 by u? and then what do you do? do you replace u by 3x^2 at the end or what?

http://s16.postimage.org/550ki472r/lol.jpg

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  • $\begingroup$ Precisely. Use familiar Taylor series for various functions and simply replace $x$ with $f(x)$. $\endgroup$
    – glebovg
    Oct 23, 2012 at 21:37

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Exactly! We have $$\ln(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+\cdots$$ when $-1 \lt t \le 1$. Set $t=3x^2$. We get the series $$3x^2-\frac{9x^4}{2}+\frac{27x^6}{3}-\frac{81x^8}{4}+\cdots.$$

Note that we will have convergence precisely if $|x|\le \dfrac{1}{\sqrt{3}}$.

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