I'm having a hard time integrating, if anyone can help, thank you
Q quen New member Joined Aug 9, 2021 Messages 1 Aug 9, 2021 #1 I'm having a hard time integrating, if anyone can help, thank you
L lookagain Elite Member Joined Aug 22, 2010 Messages 3,190 Aug 9, 2021 #2 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.
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.
D Deleted member 4993 Guest Aug 9, 2021 #3 quen said: I'm having a hard time integratingView attachment 28506, if anyone can help, thank Click to expand... This is a laborious integral. Wolfram-alpha says: Have you dealt with "hyperbolic functions"?
quen said: I'm having a hard time integratingView attachment 28506, if anyone can help, thank Click to expand... This is a laborious integral. Wolfram-alpha says: Have you dealt with "hyperbolic functions"?
L lookagain Elite Member Joined Aug 22, 2010 Messages 3,190 Aug 9, 2021 #4 quen said: I'm having a hard time integratingView attachment 28506, if anyone can help, thank you Click to expand... quen, after looking at what Subhotosh Khan posted, I feel you were given an unreasonable problem to be expected to integrate.
quen said: I'm having a hard time integratingView attachment 28506, if anyone can help, thank you Click to expand... quen, after looking at what Subhotosh Khan posted, I feel you were given an unreasonable problem to be expected to integrate.