Does the following improper integral converge or diverge $$\int\limits_{-1}^{1} \frac{\arctan{(5x+2)}}{\sqrt{1-x^2}} dx.$$
It seems that the following integral converges but I would like to find $f(x)$ such that $\int\limits_{-1}^{1} \frac{\arctan{(5x+2)}}{\sqrt{1-x^2}} dx \leq \int\limits_{-1}^{1} f(x) dx$ so it would integrate reasonably and where I could estimate that if the integral of $f(x)$ converges then the first integral converges as well.
Unfortunately I am stuck with finding the following function. If someone could help me that I would be really grateful.
Split the interval into three parts, writing the improper integral as $$\int\limits_{-1}^{1} \frac{\arctan{(5x+2)}}{\sqrt{1-x^2}} dx $$ $$=\int_{-1}^{-1/2} \frac{\arctan{(5x+2)}}{\sqrt{1-x^2}}dx+\int_{-1/2}^0 \frac{\arctan{(5x+2)}}{\sqrt{1-x^2}}dx+\int_0^1 \frac{\arctan{(5x+2)}}{\sqrt{1-x^2}}dx`.$$ The second of the three integrals on the right side above does not present any problems with respect to convergence/divergence. Let us look first at the improper integral $$\int_{-1}^{-1/2} \frac{\arctan{(5x+2)}}{\sqrt{1-x^2}}dx.$$ The numerator is between arctan(-3) and arctan(-1/2) in the interval (-1,-1/2], so the integral converges iff $$\lim_{a \to -1}\int_a^{-1/2} \frac{1}{\sqrt{1-x^2}}dx$$ exists. Upon substitution $x=\sin \theta$ the integral becomes $\lim_{a \to -1}(-\pi/6- \text {arcsin}(a))=-\pi/6+\pi/2.$ Now look at the improper integral $$\int_0^1\frac{\arctan{(5x+2)}}{\sqrt{1-x^2}}dx.$$ On the interval [0,1], the numerator is between arctan(2) and arctan(7) so the improper interval converges iff $$\lim_{b \to 1}\int_0^b \frac{1}{\sqrt{1-x^2}}dx$$ exists. Upon substitution $x=\sin \theta$ we see that the limit is $\pi/2.$ Thus the given integral converges.