how to solve and sketch $\int\frac{1}{y^{4/5}(1-y)}\,\mathrm{d}y=\,\int\mathrm{d}x$ I want to solve the following equation and then be able to sketch the curve for positive values of $x$:
\begin{align}
\int\frac{1}{y^{4/5}(1-y)}\,\mathrm{d}y&=\,\int\mathrm{d}x\\
&=\,x+c,
\end{align}
where $c$ is a constant of integration.
The thing is I don't know how to solve the integral; the wolfram engine gives a very long complicated answer which makes the sketching part quite cumbersome.
So, what would be an alternative and efficient way to solve this problem?
Is numerical integration appropriate here? Should I try to solve it using a programming language or maybe MATLAB?
Thank you in advance.
 A: Do you want to have the graph or the inverse function ?
The graph you can get as follows, there is no need for an inversion:
First sketch the curve for $\,\displaystyle y=-c+\int\frac{1}{x^{4/5}(1-x)}dx\,$ and then reflect the graph 
at the bisector of the first and third part (means: at the line $\,y=x\,$) .
Note: 
Wolfram alpha gives the approximation  
$\displaystyle\int\frac{1}{x^{4/5}(1-x)}=-1.90211\arctan(0.32492-1.05146 \sqrt[5]{x} )-$
$-0.309017\ln(\sqrt[5]{x^2}-0.618034\sqrt[5]{x}+1)+0.809017\ln(\sqrt[5]{x^2}+1.61803\sqrt[5]{x}+1)-$
$-\ln(1-\sqrt[5]{x})+1.17557\arctan(1.7013\sqrt[5]{x}+1.37638) + constant$ 
which seems to be good enough for a calculation to get the graph.
The exact solution is:
$\displaystyle\int\frac{1}{x^{4/5}(1-x)}= $
$\displaystyle-\frac{\sqrt{5}-1}{4} \ln(\sqrt[5]{x^2}-\frac{\sqrt{5}-1}{2}\sqrt[5]{x}+1)+\frac{\sqrt{5}+1}{4} \ln(\sqrt[5]{x^2}+\frac{\sqrt{5}+1}{2}\sqrt[5]{x}+1)-\ln(1-\sqrt[5]{x})$
$\displaystyle+\frac{1}{2}\sqrt{10+2\sqrt{5}} \arctan(\frac{4\sqrt[5]{x}-\sqrt{5}+1}{\sqrt{10+2\sqrt{5}}})+ \frac{1}{2}\sqrt{10-2\sqrt{5}}\arctan(\frac{4\sqrt[5]{x}+\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}})$
$+ constant$ 
A: For $y\approx0$ the differential equation looks like $y'=y^{4/5}$ which is separable and solves as $(y^{1/5})'=\frac15y^{-4/5}y'=\frac15$ or 
$$y(t)=\Bigl(y_0^{1/5}+\tfrac15(t-t_0)\Bigr)^5.$$
This means solutions move away from $0$ with increasing $t$. $y\equiv 0$ is a solution and solutions can branch out at any time, $y(t)=0$ for $t<t_0$, $y(t)=\Bigl(\frac{t-t_0}5\Bigr)^5$ for $t\ge t_0$.
For $y\approx 1$ the DE is close to $y'=1-y$ or $$y=1+(y_0-1)e^{t_0-t}.$$
This means that $y\equiv 1$ is a stable stationary solution.
In between the solutions are monotonically increasing for $y\in(0,1)$ and decreasing for $y>1$. This should be sufficient for a sketch of the solution family.

To get an exact solution use the substitution
$y=u^5$, $dy=5u^4du$ to transform
$$
\int \frac{dy}{y^{4/5}(1-u)}=\int\frac{5du}{1-u^5}
$$
This can be solved via partial fraction decomposition and leads to the observed terms of high complexity.
