I'm trying to find the antiderivative of the following function:
$$f(x)= \frac{1}{\left({1+x^2}\right)^\frac{2}{3}}$$
Please could you help me in finding how could I find the primitive of this function? Is there any particular technique concerning this types of functions?
Thanks in advance
Denoting $f(x) = \int_0^x \frac{1}{\left(1 + t^2 \right)^{\frac{2}{3}}} \, \mathrm{d}t$, by the FTC you know $f$ is an antiderivative you're looking for. Then $$ \int_0^x \frac{1}{\left(1 + t^2 \right)^{\frac{2}{3}}} \, \mathrm{d}t\overset{\color{purple}{u = \frac{t^2}{x^2}}}{=} \int_{0}^{1}\frac{1}{ \left(1 + x^2u \right)^{\frac{2}{3}}} \frac{x}{2\sqrt{u}} \mathrm{d}u =x\frac{1}{B\left(\color{green}{\frac{1}{2}}, \color{maroon}{\frac{3}{2}}- \color{green}{\frac{1}{2}}\right)}\int_{0}^{1}\frac{u^{\color{green}{\frac{1}{2}} -1}(1-u)^{\color{maroon}{\frac{3}{2}} - \color{green}{\frac{1}{2}} -1}}{ \left(1 - (\color{red}{-x^2})u \right)^{\color{blue}{\frac{2}{3}}}} \mathrm{d}u $$ since $B\left(\frac{1}{2}, \frac{3}{2}- \frac{1}{2}\right) = 2$ (with $B(x,y)$ being the Beta function). But recalling that the Hypergeometric function has integral representation $$ _2 F_1(\color{blue}{a},\color{green}{b};\color{maroon}{c};\color{red}{z}) = \frac{1}{B(\color{green}{b}, \color{maroon}{c}-\color{green}{b})} \int_0^1\frac{u^{\color{green}{b}-1}(1-u)^{\color{maroon}{c}-\color{green}{b}-1}}{(1-\color{red}{z}u)^{\color{blue}{a}}}\mathrm{d} u $$ you can conclude that $$ \boxed{\int\frac{1}{\left(1 + x^2 \right)^{\frac{2}{3}}} \, \mathrm{d}x = x\, _2F_1\left(\frac{2}{3},\frac{1}{2};\frac{3}{2};-x^2\right) +C} $$ where you recover the answer that @Martin R posted in the comments.