I want to calculate the integral $\int \frac{1+\sqrt{x+1}}{1-\sqrt[3]{x+1}}\, dx$.
We substitute $u=x+1 \Rightarrow x=u-1:=\phi (u)$. Then $\phi '(u)=1$.
So, we get \begin{equation*}\int \frac{1+\sqrt{x+1}}{1-\sqrt[3]{x+1}}\, dx=\int \frac{1+\sqrt{\phi (u)+1}}{1-\sqrt[3]{\phi (u)+1}}\cdot \phi'(u)\, du=\int \frac{1+\sqrt{u}}{1-\sqrt[3]{u}}\, du\end{equation*}
Do we substitute now $w=1-\sqrt[3]{u}\Rightarrow \sqrt[3]{u}=1-w \Rightarrow u=(1-w)^3:=\phi (w)$ ?
Then we get $\phi '(w)=3(1-w)^2\cdot (1-w)'=-3(1-w)^2$.
And so
\begin{equation*}\int \frac{1+\sqrt{u}}{1-\sqrt[3]{u}}\, du=\int \frac{1+\sqrt{\phi (w)}}{1-\sqrt[3]{\phi (w)}}\cdot \phi '(w)\, dw=\int \frac{1+\sqrt{(1-w)^3}}{1-\sqrt[3]{(1-w)^3}}\cdot (-3(1-w)^2)\, dw\end{equation*}
Is this correct so far? Or do we not apply this substition?
substitute $$x+1=t^6$$ or $$x=t^6-1$$ then we get $$dx=6t^5dt$$ and out integral will be $$\int\frac{1+t^3}{1-t^2}\cdot 6t^5dt$$ note that $$\frac{1+t^3}{1-t^2}=\frac{1-t+t^2}{t-1}$$ and write yout Integrand in the form $$\frac{t^5(t^2-t+1)}{t-1}={t}^{6}+{t}^{4}+{t}^{3}+{t}^{2}+t+1+ \left( t-1 \right) ^{-1}$$ since the degree of the polynomial in the numerator is greater than the polymial in the denominator you must do Polynom division