Find the value of undefinite integral

Question

Find $$\int \frac{dx}{(x+1)^{1/2}+(x+1)^{1/3}}$$

I have tried with let $u=(x+1)^{1/2}+(x+1)^{1/3}$ but I have nothing to solve that undefinite integral. please give me a clue for solve it.

Answer 1

$$\begin{align*} \int \frac{dx}{\left ( x+1 \right )^{1/2}+ \left ( x+1 \right )^{1/3}} &\overset{u^6=x+1}{=\! =\! =\! =\!}\int \frac{6u^5}{u^3+ u^2}\, du \\ &= \int \left ( 6u^2 - 6u - \frac{6}{u+1} + 6 \right )\, du\\ &= u\left ( 2u^2-3u+6 \right )-6\log (u+1) \end{align*}$$

Now substite $u$ with $\sqrt[6]{x+1}$ and add the constant and you are done.

Answer 2

Use the substitution $u=x+1,$ and then $s=^6\sqrt{u}$ to get

\begin{equation*} 6\int \frac{s^3}{s+1}. \end{equation*}

This is the same as

\begin{equation*} 6\int (s^2-s-\frac{1}{s+1}+1)ds. \end{equation*}

Integrate each term and re-substitute.