Evaluation of $\int_{0}^{2}\frac{\arctan(x)}{x^2+x+2}dx$

Asked May 22 '16 at 13:13

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Evaluation of $\displaystyle \int_{0}^{2}\frac{\arctan(x)}{x^2+x+2}dx$

$\bf{My\; Try::}$ Let $\displaystyle I = \int_{0}^{2}\frac{\arctan(x)}{x^2+x+2}dx = \int_{0}^{2}\frac{\arctan(x)}{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}dx$

Now Put $\displaystyle a=\frac{1}{2}$ and $\displaystyle b=\frac{\sqrt{7}}{2}\;,$ So Integral $\displaystyle I = \int_{0}^{2}\frac{\arctan x}{(x+a)^2+b^2}dx$

Now Put $x+a=b\tan \theta\;,$ Then $dx = b\sec^2 \theta d\theta$

So Integral $\displaystyle I = \frac{1}{b}\int\arctan(b\tan \theta-a)d\theta$

Now How can I solve after that, Help me

Thanks

asked May 22 '16 at 13:13

juantheron

53,015

i think it gives not a solution in the known elementary functions. – Dr. Sonnhard Graubner May 22 '16 at 13:19
The answer depends on a pretty complicated combination of $\log,\log^2,\arctan$ and $\text{Li}_2$. – Jack D'Aurizio May 22 '16 at 13:23
@juantheron: Dr Sonnhard Graubner's comment is right. – Piquito May 22 '16 at 13:24
In fact, the commentary of Jack D'Aurizio, always enlightening, says the same but with more precision. – Piquito May 22 '16 at 13:34
Today's JEE Advanced question? – N.S.JOHN May 22 '16 at 14:33
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Is there any chance that the interval of integration starts from $-1/3$ instead of $0$? If that was the case, we could have utilized the very nice transformation of the integral under $x\mapsto \frac{1-x}{1+x}$, and conclude that its closed form is $ \frac{\pi}{4 \sqrt{7}} \tan^{-1} \frac{7 \sqrt{7}}{13}$. If only... – Noam Shalev - nospoon May 22 '16 at 15:39
Try: Setting $I(a,b)=\int{\arctan(b\tan \theta -a) d\theta}$, differentiate with respect to $a$ or $b$ and see what happens, i.e., try to relate the result with well known formulas – Mohammad W. Alomari May 26 '16 at 00:54

Evaluation of $\int_{0}^{2}\frac{\arctan(x)}{x^2+x+2}dx$

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