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How can I integrate this: $$\int_1^3 \frac{\ln(x)}{\ln(x)+\ln(4-x)}dx ?$$ I tried making a substitution $t=4-x$ but it doesn't help.

Are there any tricks to solve this integral easier?

Shaun
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Lola
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    What did you get after the substitution $t=4-x$? As a quick hint, it works. (...and this is why we say you should show your work...) – Simply Beautiful Art Jun 19 '17 at 15:43
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    *sigh* I suppose I shall vote to close this question due to lack of context, as its been about 10 minutes and you still haven't shown your work. If you had, we could've pointed out any errors in your substitution $t=4-x$, but by now with all the answers below, I suppose that'd make this a completely different question... *signs again* – Simply Beautiful Art Jun 19 '17 at 15:55
  • To whomever upvoted, I cannot find any reason to upvote this question, as it is. Hovering over the upvote button, you will find that it says "This question shows research effort; it is useful and clear." However, I cannot match any of this to the question. No efforts are shown, this question does not seem useful, and it is still unclear what the OP did wrong in his/her claimed attempt at the problem. – Simply Beautiful Art Jun 19 '17 at 16:01
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    See also: https://math.stackexchange.com/questions/1073120/integral-int-12011-frac-sqrtx-sqrt2012-x-sqrtxdx/1073121#1073121 – lab bhattacharjee Jun 19 '17 at 16:14
  • I'm sorry I didn't answer I was trying to look at integrals like this one,the mistake I made was that I forgot to change the integration bounds when I substituted.I observed that I can actually use the fact that $$\int_a^b f(x)=\int_a^b f(a+b-x) $$ in order not to make any more mistakes when substituting – Lola Jun 19 '17 at 16:18
  • Took me a while to find this ^ didn't know it exists and I just saw someone posted it down below too – Lola Jun 19 '17 at 16:19
  • @Lola That identity actually comes from the substitution $x=a+b-x$. Here, that means $x=4-t$, equivalently $t=4-x$. If you had shown us your work, we could've pointed out your mistakes much sooner, which would've probably been in the interest of everyone. Whether or not you knew that identity existed is actually irrelevant, as it is equivalent to the u-substitution. – Simply Beautiful Art Jun 19 '17 at 16:21

3 Answers3

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If we call the integral $I$ and do your substitution $t=4-x$, $dt=-dx$, $1 \mapsto 3$, $3 \mapsto 1$, and we have $$ I = -\int_3^1 \frac{\log{(4-t)}}{\log{(4-t)}+\log{t}} \, dt = \int_1^3 \frac{\log{(4-x)}}{\log{(4-x)}+\log{x}} \, dx, $$ where in the last step I have used tha $\int_a^b=-\int_b^a$, and relabelled the integration variable. Now add this expression to the original expression.

Chappers
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Use $$\int_{a}^{b}f(x) \,\mathrm dx =\int_{a}^{b}f(a+b-x) \,\mathrm dx $$

$$\mathrm {I}=\int_{a}^{b}f(x) \, \mathrm dx=\int_{a}^{b}f(a+b-x) \, \mathrm dx $$ $$ \implies \mathrm {2I}=\int_{a}^{b}\big(f(x) +f(a+b-x) \big) \,\mathrm dx $$

In your case, $f(x) +f(a+b-x)=1$

$$ \implies \mathrm {2I}=\int_{a}^{b}1 \, \mathrm dx =b-a \implies \mathrm I=\frac{b-a}{2} $$

Jaideep Khare
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Using the substitution $t=4-x$ you get $$I= \int_1^3 \frac{\ln x}{\ln x + \ln (4-x)} \ \mathrm{d}x = \int_1^3 \frac{\ln (4-t)}{\ln t + \ln (4-t)} \ \mathrm{d}t$$ Now, $$2I=RHS + LHS = \int_1^3 \frac{\ln t +\ln (4-t)}{\ln t + \ln (4-t)} \ \mathrm{d}t = \int_1^3 1\ \mathrm{d}t = 2$$ Hence your integral evaluates $I=1$.

Crostul
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