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I have to find this integral:

$$\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}} dx$$

My attempt was to split the integral:

$$\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}}=\int_{\frac{1}{4}}^4\frac{x\arctan x+1}{x\sqrt{x^2+1}}+\int_{\frac{1}{4}}^4\frac{\arctan x-1}{x\sqrt{x^2+1}}$$

The first integral is:

$$\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}}=\int_{\frac{1}{4}}^4\frac{1}{x}(x\arctan x)'dx=\arctan x|_{\frac{1}{4}}^4+\int_{\frac{1}{4}}^4\frac{1}{x}\arctan x dx$$

but I am stuck here.

2 Answers2

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Hint. Let $t=1/x$ then $$\begin{align}I&:=\int_{\frac{1}{4}}^4\frac{(x+1)\arctan(x)}{x\sqrt{x^2+1}} dx= \int^{\frac{1}{4}}_4\frac{(1/t+1)\arctan(1/t)}{(1/t)\sqrt{1/t^2+1}} \frac{-dt}{t^2}\\ &=\int_{\frac{1}{4}}^4\frac{(t+1)\arctan(1/t)}{t\sqrt{t^2+1}} \,dt =\frac{\pi}{2}\int_{\frac{1}{4}}^4\frac{t+1}{t\sqrt{t^2+1}} \,dt-I. \end{align}$$ where at the last step we used the identity $\arctan(t)+\arctan(1/t)=\pi/2$ for $t>0$.

Now it should be easy to find $I$.

Robert Z
  • 145,942
4

The bounds suggest the subtitution $u=\frac{1}{x}\Rightarrow dx=-\frac{1}{u^2}du$

$$\int_{1/4}^4 \frac{(x+1)\arctan x}{x\sqrt{x^2+1}} \, dx=\int_{1/4}^4 \frac{(1+u)\arctan \frac{1}{u}}{u\sqrt{u^2+1}} \, du$$

Therefore, using $\arctan x+\arctan \frac{1}{x}=\frac{\pi}{2}$ for $x>0$, we get:

$$ \begin{aligned} \int_{1/4}^4 \frac{(x+1)\arctan x}{x\sqrt{x^2+1}} &= \frac{1}{2}\int_{1/4}^4 \frac{(x+1)(\arctan x+\arctan\frac{1}{x})}{x\sqrt{x^2+1}} \, dx\\ &= \frac{\pi}{4}\int_{1/4}^4 \frac{x+1}{x\sqrt{x^2+1}} \, dx \end{aligned} $$

Can you end it now?

LHF
  • 8,491