Solving the integral $\int \sqrt{C^2 - \frac{x^2}{1+x^2}}dx$ Let $C$ be a real constant, $C\neq0$. I am trying to solve the following integral:
$$
\int \sqrt{C^2 - \frac{x^2}{1+x^2}}dx
$$
If $C = \pm 1$, I known how to solve it (by using the substitution x=sinh(u)).

Thanks in advance.
Use the substitution $x = \tan \theta$ which after differentiating becomes $ \mathrm{d}x = \sec\theta \mathrm{d}\theta $. The integral now becomes:
$$ \int\sqrt{ C^2 - \frac{x^2}{x^2+1}}\mathrm{d}x = \int \sqrt{C^2 - \frac{\tan^2\theta}{1 + \tan^2\theta}}\sec\theta \mathrm{d}\theta = \int\frac{\sqrt{C^2-\sin^2\theta}}{\cos\theta} \mathrm{d}\theta$$
This form of the integral can be now solved but comming to the solution is rather tideous so here is a link of a site https://www.integral-calculator.com/ which will solve this integral in a step-by-step solution with only elementary functions in the final solution. If I manage to find a nicer solution I will post it.
Use the substitution $x = \tan \theta$ which after differentiating becomes $ \mathrm{d}x = \sec\theta \mathrm{d}\theta $. The integral now becomes:
$$ \int\sqrt{ C^2 - \frac{x^2}{x^2+1}}\mathrm{d}x = \int \sqrt{C^2 - \frac{\tan^2\theta}{1 + \tan^2\theta}}\sec\theta \mathrm{d}\theta = \int\frac{\sqrt{C^2-\sin^2\theta}}{\cos\theta} \mathrm{d}\theta$$
This form of the integral can be now solved but comming to the solution is rather tideous so here is a link of a site https://www.integral-calculator.com/ which will solve this integral in a step-by-step solution with only elementary functions in the final solution. If I manage to find a nicer solution I will post it.