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How to evaluate the definite integral?

$$\int \frac{7}{3x+1}dx$$ I am having difficulties to finish the question: Below is what I did: $$ =\left.\frac{7}{3}\ln|3x+1|\right|_0^4$$ $$=\frac{7}{3}\ln(\dots.$$

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  • $\begingroup$ Please learn to use mathjax to format your mathematics more legibly. $\endgroup$ – Ataraxia Jul 15 '13 at 20:09
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    $\begingroup$ You're almost there...! $\endgroup$ – DonAntonio Jul 15 '13 at 20:09
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Your integration is just fine:

All you have left to do is evaluate $$\dfrac 73 \ln|3x + 1| \Big|_0^4 = \dfrac 73 (\ln(13) - \underbrace{\ln(1)}_{\large = 0}) = \quad \frac 73 \ln(13) \quad \approx \quad5.985$$

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  • $\begingroup$ I came with 1.536 which is wrong? $\endgroup$ – Sara Sharp Jul 15 '13 at 20:09
  • $\begingroup$ not correct 1.381 $\endgroup$ – Sara Sharp Jul 15 '13 at 20:15
  • $\begingroup$ woops, I got 5.985 (I need to refresh my use of the calculator). Never worry about leaving answers as "exact", which is what you have with either of the two forms above. $\endgroup$ – Namaste Jul 15 '13 at 20:17
  • $\begingroup$ thats correct thanks!!!. Can you tell me how you come to ln(13)? $\endgroup$ – Sara Sharp Jul 15 '13 at 20:21
  • $\begingroup$ Yes, we evaluate the integral at $x = 13$ and subtract its evaluation at $x = 0$. $x = 4 \implies \frac 73\ln(3x + 1) = \frac 73 \ln(3\cdot 4 + 1) = \frac 73 \ln(13)$. At x = 0, we have $\frac 73\ln(3\cdot 0 + 1) = \frac 73\ln(1) = 0$ $\endgroup$ – Namaste Jul 15 '13 at 20:23
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Hint:take $u=3x+1\to du=3dx$$$\int_0^4\frac{7}{3x+1}dx=\frac73\int_0^4\frac{3dx}{3x+1}=\frac73\int_0^4\frac{du}{u}=lnu|_1^{13}=\frac73(\ln(13)-ln1)=\frac73(\ln(13)=\frac13\log_e {13^7}$$

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Let's see if I guessed correctly what you wrote (use LaTeX for mathematics in this site, please!):

$$\int\limits_0^4\frac7{3x+1}dx=\frac73\int\limits_0^4\frac{3\,dx}{3x+1}=\left.\frac73\log(3x+1)\right|_0^4=\frac73\left(\log13-\log1\right)=\ldots$$

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  • $\begingroup$ I came with y= 1.536 which is wrong. Can you tell me your answer? $\endgroup$ – Sara Sharp Jul 15 '13 at 20:18
  • $\begingroup$ You have it there: $$\frac73\log 13\cong 5.9849$$ but I think it looks nicer as the left side... Also, why do you put that $\;y\;$ there?? $\endgroup$ – DonAntonio Jul 15 '13 at 20:20
  • $\begingroup$ when I use my calculator (7)/(3)ln(13) I'm getting this answer: (7ln(13)/(3) $\endgroup$ – Sara Sharp Jul 15 '13 at 20:24
  • $\begingroup$ And that's correct, @SaraSharp .I'm afraid you don't quite know how to use that calculator, and perhaps using some parentheses and operations order would help. Try " log(13) * 7 =" , and then divide the result by three... $\endgroup$ – DonAntonio Jul 15 '13 at 20:26

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