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Krizalid · #5 December 7th, 2007, 07:29 AM

Nice problem

Quote:

Originally Posted by liyi

$\int_0^1\frac1{(x^2-x+1)(e^{2x-1}+1)}\,dx$

Define $u=2x-1,$

$\int_0^1 {\frac{1}{{\left( {x^2 - x + 1} \right)\left( {e^{2x - 1} + 1} \right)}}\,dx} = \int_{ - 1}^1 {\frac{2}{{\left( {u^2 + 3} \right)\left( {e^u + 1} \right)}}\,du} .$

Substitute $u=-\varphi,$

$\int_{ - 1}^1 {\frac{2}{{\left( {u^2 + 3} \right)\left( {e^u + 1} \right)}}\,du} = \int_{ - 1}^1 {\frac{2}{{\left( {\varphi ^2 + 3} \right)\left( {e^{ - \varphi } + 1} \right)}}\,d\varphi ,}$ so

$\tau = \int_{ - 1}^1 {\frac{2}{{\left( {u^2 + 3} \right)\left( {e^u + 1} \right)}}\,du} = \int_{ - 1}^1 {\frac{2}{{\left( {u^2 + 3} \right)\left( {e^{ - u} + 1} \right)}}\,du} .$

And finally

$2\tau = \int_{ - 1}^1 {\frac{2}{{u^2 + 3}}\,du} = \frac{{2\pi }}{{3\sqrt 3 }}\,\therefore \,\tau = \frac{\pi }{{3\sqrt 3 }}.$

The following users thank Krizalid for this useful post:
liyi
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