Integral: int [ 1/(x - x^3/5 ] dx

[kase11]

Can someone please help me integrate this:

......1......
x-x^3/5

or

1/(x-x^3/5)

Thanks a lot

 

[pka]

In the first place look at some algebra:
\(\displaystyle \L \frac{1}{{x - x^{3/5} }} = \frac{{x^{ - 3/5} }}{{x^{2/5} - 1}}\).

Find the derivative of \(\displaystyle \L y = \ln \left[ {x^{2/5} - 1} \right]\).

Now can you finish?

 

[Deleted member 4993]

substitute

u = x^(1/5)

du = 1/5 * x^(-4/5) dx

dx = 5u^(4) du

dx/(x-x^3/5)

= 5u^(4) du/(u^5 - u^3)

= 5u du/(u^2 - 1)

Integrate and back-substitute

 

[galactus]

This is close to SK's, but I will go ahead.

You have:

\(\displaystyle \L\\\int\frac{1}{x-x^{\frac{3}{5}}}dx\)

Factor and get:

\(\displaystyle \L\\\int\frac{1}{x^{\frac{3}{5}}(x^{\frac{2}{5}}-1)}dx\)

Now let \(\displaystyle \L\\u=x^{\frac{2}{5}}-1, \;\ du=\frac{2}{5x^{\frac{3}{5}}}dx, \;\ \frac{5}{2}du=\frac{1}{x^{\frac{3}{5}}}dx\)

Then you have:

\(\displaystyle \L\\\frac{5}{2}\int\frac{1}{u}du\)

Hey, I just noticed, this is number 3000

 

[kase11]

Thanks a lot to everyone