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$$\int_{1}^2 \log (t^2+t+1)\mathrm dt$$

I 'd like to calculate the above value.

my question is,

1) how to integrate $\log x$?

1-1) is there any way to calculate above value without integration?

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    $\begingroup$ try to integrate by parts $\endgroup$ – Mosk Jun 11 '17 at 11:06
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    $\begingroup$ Ad 1): that won't help you much. 1-1) Yes, sometimes, many prefer to ask at Stack Exchange instead of calculating it themselves, but few get help without any own effort. You may want to look up "integration by parts". $\endgroup$ – Professor Vector Jun 11 '17 at 11:15
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    $\begingroup$ A) Integrate by parts. B) Complete the square $(t+1/2)^2=\cdots$. $\endgroup$ – Jyrki Lahtonen Jun 11 '17 at 11:20
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$$\begin{align} \int_1^2 \log(t^2 + t + 1) dt &= \left[t\log(t^2 +t+1)\right]_1^2 - \int_1^2 t\cdot\frac{2t+1}{t^2+t+1}dt \\ &= 2\log7-\log3-\int_1^2 \frac{2t^2+t}{t^2+t+1}dt \\ &= 2\log7-\log3-\int_1^2 \frac{2t^2+2t+2-\frac{1}{2}(2t+1)-\frac{3}{2}}{t^2+t+1} dt\\ &= 2\log7-\log3-\int_1^2 2dt+\frac{1}{2} \int_1^2\frac{2t+1}{t^2+t+1}dt + \frac{3}{2}\int_1^2 \frac{1}{t^2+t+1} dt \\ \end{align}$$ These integrals can now all be dealt with separately, using other standard techniques. It looks to me like integration by parts is the most efficient way of solving this problem.

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Answer of your first question is 'By parts'.the procedure is , taking log(t²+t+1) as first function,and 1 i.e. t^0 treated as second function. Now answer of your second one is,no,in my point of view.

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  • $\begingroup$ Aha. We can think of 1 equal to t^0 thx $\endgroup$ – Daschin Jun 11 '17 at 11:19
  • $\begingroup$ Please telling me how can i type power of t be 0, $\endgroup$ – Sital Pal Jun 11 '17 at 11:25
  • $\begingroup$ Use $ around the expression, like this $t^0$ $\endgroup$ – B. Mehta Jun 11 '17 at 11:26

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