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$$\int_0^e\frac{dx}{x(\ln^2{x}+2\ln x+5)}=|\text{subs}:\; t=\ln x; x= e^t; dx = e^t dt|$$ $$ =\int_\infty^1 \frac{dt}{t^2+2t+5} = \int_\infty^1\frac{dt}{(t+1)^2+4} $$ $$= \left[{\frac12\arctan\frac{t+1}2}\right]_\infty^1\; =\;\; \frac{\pi}{8}-\frac12\arctan\left(\lim_{t\to0} \frac{t+1}2\right)\; = \;\;\frac{\pi}8 - \frac{\pi}4 \;= \;-\frac{\pi}8$$

The result isn't right, but I can't see my mistakes. Do you see where did I do wrong?

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It should be $\int^1_{-\infty}$. –  anon Nov 26 '12 at 22:15
    
You are right! Thank you :) –  user50222 Nov 26 '12 at 22:22
    
I was quite puzzled by "$(t+1)^24$", but then I realized what must have been intended and changed it to $(t+1)^2+4$. –  Michael Hardy Nov 27 '12 at 0:43
    
I'd write $(\ln x)^2$, since $\ln^2 x$ could be construed as $\ln\ln x$. But maybe I'd write $(\log x)^2$, or if that upsets some people, $(\log_e x)^2$. –  Michael Hardy Nov 27 '12 at 0:44

1 Answer 1

up vote 2 down vote accepted

Tthe lower limit in $t$ should be $-\infty$ and you lost a $+$ before the $4$ in the denominator. The lower limit should be evaluated as $t \to -\infty$, which gives an arctan of $-\frac \pi 2$ and the last term has the wrong sign, giving $\frac {3\pi}8$, confirmed by Alpha

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(I think the missing $+$ is just a typo.) –  anon Nov 26 '12 at 22:20

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