EclipseEnigma wrote:Spoiler:
Although the answer should clearly be log(log x)
I think you made a sign error in that.
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EclipseEnigma wrote:Spoiler:
Although the answer should clearly be log(log x)
t0rajir0u wrote:What does the graph of [imath]y^5 = 1 - x^7[/imath] look like? How does it relate to the graph of [imath]y^7 = 1 - x^5[/imath]?
csrjjsmp wrote:One of my favorites:
[math]\int_0^1(1-x^7)^{1/5} - (1-x^5)^{1/7}dx[/math]
Excalibur0998 wrote:t0rajir0u wrote:What does the graph of [imath]y^5 = 1 - x^7[/imath] look like? How does it relate to the graph of [imath]y^7 = 1 - x^5[/imath]?
I put them into wolfram alpha, and they look very similar...but what does [imath]y^5 = 1 - x^7[/imath] have to do with [imath](1 - x)^7[/imath]? Anyway, I finally sort of gave up on this integral so I went to my dad and he showed me how to do it usingbut if there's another approach (graphical approach) then I'd be interested in hearing it.Spoiler:
mr-mitch wrote:Test your brilliance:
[math]\int \sqrt{x+k \over x} dx[/math]
I will award one complex prize to the first person who solves it.
PS It's hell, although that's possibly an understatement.
mr-mitch wrote:It takes quite a while to do
mr-mitch wrote:Test your brilliance:
[math]\int \sqrt{x+k \over x} dx[/math]
I will award one complex prize to the first person who solves it.
PS It's hell, although that's possibly an understatement.
Fluid_Dynamic wrote:Here's one that I stumped my high school class with for a few days:
[imath]\int \frac{1}{x \log x} \,dx[/imath]
Try solving it through integration by parts ( [imath]u = \frac{1}{\log x}, dv = \frac{1}{x} \,dx[/imath] )
Plasma_Wolf wrote:Fluid_Dynamic wrote:Here's one that I stumped my high school class with for a few days:
[imath]\int \frac{1}{x \log x} \,dx[/imath]
Try solving it through integration by parts ( [imath]u = \frac{1}{\log x}, dv = \frac{1}{x} \,dx[/imath] )
I assume that the u has to be a v or the dv has to be an u.
The way I get to the answer:Spoiler:
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