Find $\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}} dx$ 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.
 A: 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 $I$.
A: 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?
