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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 $ln\left((x+1)^3 \cdot (2x+3)\right)$

(replace parentheses with absolute value signs, I don't know how to make those in tex).

This is not the correct answer according to my calculations and Wolfram Alpha http://www.wolframalpha.com/input/?i=integrate+%288x%2B11%29%2F%28%282x%2B3%29%28x%2B1%29%29

Any ideas what's going on?

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2  
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. –  Trancot Nov 29 '13 at 9:58
    
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) –  DetlevCM Nov 29 '13 at 12:57

3 Answers 3

up vote 4 down vote accepted

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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Yes, I see that now. Thank you. –  Paze Nov 29 '13 at 9:59
    
Hear, hear! Calculators, boo! –  Trancot Nov 29 '13 at 9:59

The answer is correct.

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

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Whoops. Looks like I was too quick to post this time. Thank you for clarifying. –  Paze Nov 29 '13 at 9:58

Looks to me like WolframAlpha agrees with the answer from NSpire.

Are you sure they disagree?

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