# Evaluate $\int {dx\over \sqrt[6]{(x-7)^7(x-5)^5}}$

Evaluate : $$\int {dx\over \sqrt[6]{(x-7)^7(x-5)^5}}$$

The most obvious idea would be setting: $$t^6 = (x-7)^7(x-5)^5$$

But then the problem becomes too complicated, since finding $$dx$$ is not that simple. I believe there must be some smart substitution which I couldn't find.

I was then thinking of setting: $$t = x-7\\ dx = dt\\ \sqrt[6]{(x-7)^7(x-5)^5} = \sqrt[6]{t^7(t+2)^5}$$ So that the integral becomes: $$\int {dt\over \sqrt[6]{t^7(t+2)^5}}$$

Which again leads to complicated calculations. Could someone suggest a way to solve that integral?

## 1 Answer

Let's do a more general case: $$I(n)=\int \frac{dx}{\sqrt[n]{(x-(n+1))^{n+1}(x-(n-1))^{n-1}}}\overset{x-n=t}=\int \frac{dt}{\sqrt[n]{(t-1)^{n+1}(t+1)^{n-1}}}$$ $$=\int \frac{dt}{(t^2-1)\sqrt[n]{ \frac{t-1}{t+1}}}\overset{(t-1)/(t+1)=y}=\frac12\int \frac{1}{y}\frac{dy}{\sqrt[n]{y}}=-\frac{n}{2}\frac{1}{\sqrt[n]{y}}+C,\quad y=\frac{x-n-1}{x-n+1}$$

• Whoa, that's neat, thank you! – roman Oct 17 '19 at 17:25
• It should be said that the domain of validity of such an integral is outside interval $[n-1,n+1]$. – Jean Marie Oct 18 '19 at 8:29