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Evaluate $$I=\int_2^6 \frac{\ln(x-1)}{x^2+2x+2}dx$$

I tried the substitution $t=x-1$ which gave $$I=\int_1^5 \frac{\ln(t)}{t^2+4t+5}dt$$ Then I let $u=\ln t$ which gave $$I=\int_0^{\ln 5} \frac{ue^u}{e^{2u}+4e^u+5}du$$ I don't see other useful substitutions and I don't know how to finish

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Symmetry works nice in this case. $$I=\int_2^6 \frac{\ln(x-1)}{x^2+2x+2}dx\,\overset{x-2\rightarrow x}=\,\int_0^4 \frac{\color{red}{\ln(1+x)}}{x^2+6x+10}dx$$ With a substitution of $$x=\frac{4-y}{1+y}\Rightarrow dx= - \frac{5}{(1+y)^2}dy$$ $$\require{cancel} \Rightarrow I= \int_0^4 \ln\left(1+\frac{4-y}{1+y}\right)\frac{\cancel{(1+y)^2}}{\cancel 5(y^2+6y+10)}\frac{\cancel 5}{\cancel{(1+y)^2}}dy\overset{y=x}=\int_0^4 \frac{\color{blue}{\ln\left(\frac{5}{1+x}\right)}}{x^2+6x +10}dx $$Adding this result with the previous integral gives: $$\require{cancel} 2I=\int_0^4 \frac{\cancel{\color{red}{\ln(1+x)}}+\color{blue}{\ln 5-\cancel{\ln(1+x)}}}{x^2+6x+10}dx$$ $$\Rightarrow I=\frac{\ln 5}{2} \int_0^4 \frac{1}{x^2+6x+10}dx=\frac{\ln 5}{2} \arctan\left(\frac2{11}\right)$$


One can of course do directly $\displaystyle{x=\frac{t+4}{t-1}}$ when the bounds are $2$ and $6$ to arrive at the same result by the same method.

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    $\begingroup$ Thank you! How did you find the substitution $x= \frac{4-y}{1+y}$? $\endgroup$ – sgc Feb 22 '19 at 16:20
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    $\begingroup$ A while ago I was passionate by this type of integrals. You can find here in my answer a subtitution for the general result: math.stackexchange.com/a/3003910/515527. $\endgroup$ – Zacky Feb 22 '19 at 16:24
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    $\begingroup$ I don't have any intuition of why to find it at first glance, I just tried many combinations to get the right one (it took me more than a month). This might be in your interest too: math.stackexchange.com/q/3003880/515527 and this: math.stackexchange.com/a/3037342/515527 $\endgroup$ – Zacky Feb 22 '19 at 16:26
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    $\begingroup$ Wow, thank you so much for everything! $\endgroup$ – sgc Feb 22 '19 at 16:28
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\begin{align}I=\int_2^6 \frac{\ln(x-1)}{x^2+2x+2}dx\end{align}

1) Clean up the logarithm, Perform the change of variable $y=x-1$,

\begin{align}I=\int_1^5 \frac{\ln x}{x^2+4x+5}dx\end{align}

2) One wants to obtain a symmetric polynomial (i.e: $ax^2+bx+a$). Perform the change of variable $x=\sqrt{5}u$,

\begin{align}I&=\sqrt{5}\int_{\frac{1}{\sqrt{5}}}^{\frac{5}{\sqrt{5}}} \frac{\ln \left(u\sqrt{5}\right)}{5u^2+4\sqrt{5}u+5}du\\ &=\sqrt{5}\int_{\frac{1}{\sqrt{5}}}^{\sqrt{5}} \frac{\ln u}{5u^2+4\sqrt{5}u+5}du+\sqrt{5}\ln\left( \sqrt{5}\right)\int_{\frac{1}{\sqrt{5}}}^{\sqrt{5}} \frac{1}{5u^2+4\sqrt{5}u+5}du\\ \end{align}

The first integral is $0$ (perform the change of variable $y=\frac{1}{u}$)

\begin{align}I&=\ln\left( \sqrt{5}\right)\left[\arctan\left(\frac{10u+4\sqrt{5}}{2\sqrt{5}}\right)\right]_{\frac{1}{\sqrt{5}}}^{\sqrt{5}}\\ &=\frac{1}{2}\ln 5\left(\arctan 7-\arctan 3\right)\\ &=\boxed{\frac{1}{2}\arctan\left(\frac{2}{11}\right)\ln 5} \end{align}

NB:

1) a) It's often a good idea to clean up the logarithm.

I mean one perform a change of variable to obtain that the logarithm argument be $x$.

b) Every change of variable $y=\frac{a}{x}$, or $y=ax$, doesn't make "dirty" the logarithm, since,

\begin{align} \ln (ax)&=\ln a+\ln x\\ \ln\left(\frac{a}{x}\right)&=\ln a-\ln x\end{align}

2) Why do we want to obtain symmetric polynomial?

Because, if $\alpha>1$,$a,b,c$ real,

\begin{align}\int_{\frac{1}{\alpha}}^\alpha\frac{\ln x}{ax^2+bx+a}\,dx=0\end{align}

(i assume that the polynomial doesn't have a root in $\left[\frac{1}{\alpha},\alpha\right]$)

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    $\begingroup$ Well done. +1 for the explanations. $\endgroup$ – Allawonder Feb 23 '19 at 2:33

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