<s>[INST] Integrating $\frac{\arctan x}{x\sqrt{\smash[b]{1-x^2}}}$ How to evaluate
$$\int_0^1\frac{\arctan x}{x\sqrt{1-x^2}}\,\mathrm dx\text{?}$$
The steps I can think of is integration by parts as
$$\int_0^1\frac{\arctan x}{x\sqrt{1-x^2}}\,\mathrm dx=\int_0^1\frac{\arctan x}{x}\,\mathrm d(\arcsin x)$$
or integration by substitution using $x=\sin t$ as
$$\int_0^1 \frac{\arctan x}{x\sqrt{1-x^2}}\,\mathrm dx = \int_0^{\frac{\pi}{2}}\frac{\arctan(\sin t)}{\sin t}\,\mathrm dt,$$
but both seems to make the problem more complicated.
Thanks a lot.
 [/INST] Let $\displaystyle I(a)=\int^{1}_{0}\frac{\tan^{-1}(ax)}{x\sqrt{1-x^2}}dx$
Now $\displaystyle I'(a)=\int^{1}_{0}\frac{1}{(1+a^2x^2)\sqrt{1-x^2}}dx$
Put $x=\tan t$. Then $dx=dt$ and changing limits
So $$I'(a)=\int^{\frac{\pi}{2}}_{0}\frac{1}{1+a^2\sin^2 t}dt=\frac{1}{1+a^2}\int^{\frac{\pi}{2}}_{0}\frac{\sec^2 t}{k^2+\tan^2 t}dt$$
Where $\displaystyle k^2=\frac{1}{1+a^2}$
So $\displaystyle I'(a)=\frac{1}{(1+a^2)\cdot k}\tan^{-1}\bigg(\frac{t}{k}\bigg)\bigg|^{\infty}_{0}=\frac{\pi}{2}\cdot \frac{1}{\sqrt{1+a^2}}$
So $$I(a)=\frac{\pi}{2}\ln\bigg|a+\sqrt{1+a^2}\bigg|$$
put $a=1$. Then $$I(1)=\frac{\pi}{2}\ln\bigg|1+\sqrt{2}\bigg|.$$
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