Question

Hello.

I'm stuck on an integral , which seems a bit difficult...

I've got a link of wolframalpha where I followed the instructions .... but It seems something I don't understand...

http://www.wolframalpha.com/input/?i=%28%28x%5E%280.5%29-1%29%5E3%29%2F%28x%29

I'll attach a photo too !

Answer 1

Picture is in high resolution...

Answer 2

Maybe du substitution has something to do with it.. but I need to know how ! :(

Answer 3

expand..simplify...integrate

Answer 4

actually in wolframa there's a next subtitution we have to make... (before doing it..) but how does number 2 appear there? ... that's what I don't get !

Answer 5

@Zarkon I thought of expanding too, but its tedious,
Leu u^2=x
2udu=dx

\[\int\limits_{}^{}\frac{(u-1)^3}{u^2}(2u)du\]

Simplify it

Answer 6

ok ...

Answer 7

thank you

Answer 8

\[\LARGE du=\frac{1}{2\sqrt{x}}dx \quad \quad ,\quad \sqrt x=u\]

then...

\[\LARGE du=\frac{1}{2u}dx \Longrightarrow 2u\cdot du=dx\]

.. I got it ! Thank you !

Answer 9

dang that is a lot of extra to ing and fro ing for not much reward

\[(a+b)^3=a^3+3a^2b+3ab^2+b^3\] you only have 4 terms
\[(\sqrt{x}-1)^3=x^{\frac{3}{2}}-3x++3\sqrt{x}-1\] divide by x and it is an exercise in adding and subtracting exponents

Answer 10

i would find a substitution much more "tedious' but i guess it is a matter of taste