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Evaluate the definite integral. $$\int_3^7 \ln(x^{43})\,dx$$

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Is it $(ln x)^{43}$ or $ln (x^{43})$? –  Gautam Shenoy Jan 28 '13 at 17:28
    
@AndréNicolas he said "It should be $\ln(x^{43})$" –  CBenni Jan 28 '13 at 17:29
    
It is presents exactly like that in the question but I'm interpreting it as ln(x^43) –  Gabrielle Jan 28 '13 at 17:30

5 Answers 5

Recall the properties of logarithms, in particular:

$$\;\ln(a^b) = b \ln a\;\implies \;\int_3^7 \; \ln({x^{43}})\,dx \;= \; \int_3^7 43\ln{x}\,dx\; = \;43\int_3^7 \ln x\,dx$$ Then use the fact that $\;\;\displaystyle43\int_3^7\,\ln x\,dx\;=\; 43[x\ln x-x]\Big|_3^7 \;$,

and evaluate, given your bounds of integration!

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+1 I think there is needed a $dx$. –  Babak S. Jan 28 '13 at 19:40
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Thanks, @Babak: yes, I missed the $dx$ in the last integral! Thanks for catching it, and feel free to edit any post of mine when you see a typo or neglected term! –  amWhy Jan 28 '13 at 19:59

hint: $\int_3^7 \ln x^{43} \, dx=\int_3^7 43 \ln x\,dx$ and compute following integral on [3,7] $$\int \ln x =x\ln x-x$$

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This is sort of a joke answer (but correct).

If we do not know any "log laws," let $u=\log(x^{43})$, $dv=dx$. Then $du=\dfrac{43x^{42}}{x^{43}}\,dx=\dfrac{43}{x}\,dx$, and we can take $v=x$. Thus our indefinite integral is $$x\log(x^{43})-\int \frac{43}{x} x\,dx.$$

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HINT: $\ln(x^a)=a\ln(x)$ This should do the trick if you really wanted $\ln(x^{43})$

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$$\begin{align} \int_3^7 dx \: \log{x^{43}} &= 43 \int_3^7 dx \: \log{x}\\ &= 43 [(x \log{x} - x)]_3^7 \\ &= 43 (7 \log{7} - 7 - (3 \log{3} - 3)) \end{align}$$

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Thank you so much! –  Gabrielle Jan 28 '13 at 17:35
    
You're welcome. If you found an answer useful, please remember to accept it. –  Ron Gordon Jan 28 '13 at 19:34

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