How to find the integral $\int_0^1 \frac{x^{1/2}}{1+x^{1/3}}dx$? How to find $$\int_0^1 \frac{x^{1/2}}{1+x^{1/3}}dx$$ ?
My attempt: I made $t^6=x$ and got $\displaystyle \int_0^1 \frac{6t^8}{1+t^2}dt$ and got stuck.
 A: Hint: use long division:
$$\frac{t^8}{t^2+1}= t^6-t^4+t^2-1+\frac{1}{t^2+1}.$$
A: $$t^8 = (t^2)^4 = ((t^2+1)-1)^4 = \sum_{k=0}^{4}\binom{4}{k}(-1)^k (t^2+1)^{k} $$
leads to:
$$ \int_{0}^{1}\frac{t^8}{1+t^2}\,dt = \frac{\pi}{4}+\sum_{k=1}^{4}\binom{4}{k}(-1)^k \int_{0}^{1}(1+t^2)^{k-1}\,dt =\color{red}{-\frac{76}{105}+\frac{\pi }{4}}.$$
A: Hint:
$$\frac{t^8}{1+t^2} = t^6 \frac{t^2}{1+t^2} = t^6-\frac{t^6}{1+t^2}. $$
A: We can evaluate the integral by expanding the denominator in a geometric series.  Proceeding we have
$$\begin{align}
\int_0^1 \frac{t^8}{1+t^2}\,dt&=\sum_{n=0}^\infty (-1)^n\int_0^1 t^{8+2n}\,dt\\\\
&=\sum_{n=0}^\infty \frac{(-1)^n}{2n+9}\\\\
&=\sum_{n=0}^\infty\frac{(-1)^n}{2(n+4)+1}\\\\
&=\underbrace{\color{blue}{\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}}}_{\text{Series for}\,\,\arctan(1)}-\color{red}{\sum_{n=0}^3\frac{(-1)^n}{2n+1}}\\\\
&=\color{blue}{\frac{\pi}{4}}-\color{red}{\frac{76}{105}}
\end{align}$$
where we recognized the series representation for the arctangent evaluated at $1$, $\arctan(1)=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}$.
A: Try writing the numerator as $$6t^8 + 6t^6 - 6t^6 - 6t^4 + 6t^4 + 6t^2 - 6t^2 - 6 + 6.$$
A: Hint write it as $6t^8=6t^8+6t^6-6t^6$ then by dividing it by denominator you get it as $6-\frac {6t^6}{1+t^2} $
 Now write the next part of $6t^6=6t^6-6t^4+6t^4$...and continue till $t^2$ to solve an easy integrable function.
