After learning the integration of various functions with $x^2$ involved, I was given the following integration, as a challenge: $$\sqrt{1+x^3}$$
I tried various methods - too long to even try and post here. For example, I tried to integrate by parts - by assuming the second function to be $1$, I tried trigonometric substitution, I tried algebraic substitution. However, I always got stuck at some point. What would be the possible approach to integrating such a function?
For any real number of $x$ ,
When $|x|\leq1$ ,
$\int\sqrt{1+x^3}~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{3n}}{4^n(n!)^2(1-2n)}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{3n+1}}{4^n(n!)^2(1-2n)(3n+1)}+C$
When $|x|\geq1$ ,
$\int\sqrt{1+x^3}~dx$
$=\int x^\frac{3}{2}\sqrt{1+\dfrac{1}{x^3}}~dx$
$=\int x^\frac{3}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(1-2n)x^{3n}}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{\frac{3}{2}-3n}}{4^n(n!)^2(1-2n)}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{\frac{5}{2}-3n}}{4^n(n!)^2(1-2n)\left(\dfrac{5}{2}-3n\right)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{2^{2n-1}(n!)^2(2n-1)(6n-5)x^{3n-\frac{5}{2}}}+C$