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In there any way to calculate Integral

$$\int_{\frac{1}{2014}}^{2014} \frac{\tan^{-1}x}{x} dx$$

by using simple properties of definite integration. I used substitution $\tan^{-1}x=t $ and applied by parts but then I am stuck at $\int\int \csc(2t)dt$ How should I proceed? What is the most appropriate method to solve this question?

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  • $\begingroup$ Hint: Try by parts first, so you have $\int {1 \over x(x^2+1)} $ and then use partial fractions. $\endgroup$ – Morgormir Feb 18 '16 at 19:23
  • $\begingroup$ @Morgormir Your integration by parts is incorrect. You should instead get $\int\frac{\ln{x}}{1+x^{2}}$, which isn't any easier. An elementary antiderivative for this function does not exist. $\endgroup$ – David H Feb 18 '16 at 19:30
  • $\begingroup$ You are correct, I wrote out the integral with $x^2$ as it's fraction, my mistake. $\endgroup$ – Morgormir Feb 18 '16 at 20:49
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For $x>0$, $$\tan^{-1}(x)=\cot^{-1}\left(\frac1x\right)$$ Here, $$I=\int_{\frac{1}{2014}}^{2014} \frac{\tan^{-1}x}{x} dx$$ Let $x=\frac1y$. Then, $$\int_{\frac{1}{2014}}^{2014} \frac{\tan^{-1}x}{x} dx=\int^{\frac{1}{2014}}_{2014} \frac{\tan^{-1}\frac1y}{\frac1y}\left(-\frac1{y^2}\right)dy=-\int^{\frac{1}{2014}}_{2014} \frac{\tan^{-1}\frac1y}{y}dy=-\int^{\frac{1}{2014}}_{2014} \frac{\cot^{-1}y}{y}dy=-\int^{\frac{1}{2014}}_{2014} \frac{\frac{\pi}{2}-\tan^{-1}y}{y}dy=\int^{\frac{1}{2014}}_{2014} \frac{\tan^{-1}y-\frac{\pi}{2}}{y}dy=-I-\int^{\frac{1}{2014}}_{2014} \frac{\frac{\pi}{2}}{y}dy$$ Thus, $$I=\int_{\frac{1}{2014}}^{2014} \frac{\pi}{4y}dy$$ which can be easily solved.

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A straightforward one: use the substitution $u=1/x$ and the formula: $$\forall x\in\mathbb{R}_+^*,\ \arctan\left(\frac1x\right)=\frac\pi2-\arctan(x).$$ For the sake of generality, we'll compute the integral from $1/a$ to $a$ for some $a\in\mathbb{R}_+^*$: $$I=\int_{1/a}^{a}\frac{\arctan(x)}{x}\,\mathrm{d}x=\int_a^{1/a}u\arctan(1/u)\left(-\frac{\mathrm{d}u}{u^2}\right)=\int_{1/a}^a\frac{\pi/2-\arctan(u)}{u}\,\mathrm{d}u=\int_{1/a}^a\frac{\pi}{2u}\,\mathrm{d}u-I.$$ Hence $$2I=\int_{1/a}^a\frac{\pi}{2u}\,\mathrm{d}u,$$ i.e., $$I=\frac\pi4\int_{1/a}^a\frac{\mathrm{d}u}u=\frac\pi4\bigl(\ln(a)-\ln(1/a)\bigr)=\frac\pi2\ln(a).$$

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