Integrate
- Thread starter quen
- Start date
You could rewrite the integrand as \(\displaystyle \ (x^{-2/3} + 1)x dx.\)
I would recommend this u-substitution:
Let \(\displaystyle \ u = x^{-2/3}.\)
Then \(\displaystyle \ u^3 = x^{-2}.\)
And \(\displaystyle \ u^{-3} = x^{2}.\)
So, \(\displaystyle \ -3u^{-4}du = 2x dx.\)
Then continue and write your work in a post here.
I would recommend this u-substitution:
Let \(\displaystyle \ u = x^{-2/3}.\)
Then \(\displaystyle \ u^3 = x^{-2}.\)
And \(\displaystyle \ u^{-3} = x^{2}.\)
So, \(\displaystyle \ -3u^{-4}du = 2x dx.\)
Then continue and write your work in a post here.
D
Deleted member 4993
Guest
This is a laborious integral. Wolfram-alpha says:
Have you dealt with "hyperbolic functions"?
quen, after looking at what Subhotosh Khan posted, I feel you were given an
unreasonable problem to be expected to integrate.