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

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

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?

• try to integrate by parts – Mosk Jun 11 '17 at 11:06
• 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". – Professor Vector Jun 11 '17 at 11:15
• A) Integrate by parts. B) Complete the square $(t+1/2)^2=\cdots$. – Jyrki Lahtonen Jun 11 '17 at 11:20

\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.
• Use $ around the expression, like this $t^0\$ – B. Mehta Jun 11 '17 at 11:26