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Consider the integral: $$\int \frac{8x+11}{(2x+3)(x+1)}$$ My Nspire CAS tells me that the answer to this is $$\ln\left\lvert(x+1)^3 \cdot (2x+3)\right\lvert$$

This is not the correct answer according to my calculations and Wolfram Alpha

Any ideas what's going on?

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    $\begingroup$ Absolute value is ridiculous in LaTeX. It's \lvert expression \rvert. Of course, you could do any number of things, but purists always prefer this method. $\endgroup$
    – Trancot
    Nov 29, 2013 at 9:58
  • $\begingroup$ Wolfram alpha tells you the same - if you check the footer, it defines log as the natural logarithm, i.e. ln. (Still a funny way to write it) Then: 3log(x+1) + log(2x+3) = log((x+1)^3) + log(2x+3) = log((x+1)^3(2x+3) $\endgroup$
    – DetlevCM
    Nov 29, 2013 at 12:57

3 Answers 3

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HINT:

Using Partial Fraction Decomposition

$$\frac{8x+11}{(2x+3)(x+1)}=\frac A{2x+3}+\frac B{x+1}$$

Do you know $\displaystyle \ln a+\ln b=\ln ab$ and consequently $\displaystyle c\cdot\ln a=\ln (a^c)$ (assuming if each logarithm is defined) ?

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  • $\begingroup$ Yes, I see that now. Thank you. $\endgroup$
    – Paze
    Nov 29, 2013 at 9:59
  • $\begingroup$ Hear, hear! Calculators, boo! $\endgroup$
    – Trancot
    Nov 29, 2013 at 9:59
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The answer is correct.

$$\log ((x+1)^3 (2x+3)) = 3\log(x+1) + \log(2x+3)$$

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  • $\begingroup$ Whoops. Looks like I was too quick to post this time. Thank you for clarifying. $\endgroup$
    – Paze
    Nov 29, 2013 at 9:58
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Looks to me like WolframAlpha agrees with the answer from NSpire.

Are you sure they disagree?

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