anonymous
  • anonymous
Integration please. I have to use algebraic substitution with this question but I really need help. I will type it below.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[integrate \sqrt4 / \sqrt x^3 + 4\]
anonymous
  • anonymous
If I am correct I let u = x^3 +4 du/dx=3x^2 and dx=dx+du/3x^2
anonymous
  • anonymous
\[\int\frac{\sqrt 4}{\sqrt{x^3}}\,dx\]?

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anonymous
  • anonymous
Is the 3 in the bottom inside or outside the \sqrt?
Callisto
  • Callisto
Is your question :|dw:1332496172685:dw| ?
anonymous
  • anonymous
then I typed \[integrate x^1/2 (2)^-1/2 (du/3x^2) I don't know what \to do next.\]
anonymous
  • anonymous
the 4 on the bottom is NOT part of the square root.
anonymous
  • anonymous
Teacer said we HAVE to use algebraic substitution to get rid of the square roots.
Callisto
  • Callisto
So the question is|dw:1332496271369:dw|
anonymous
  • anonymous
Yes Calisto that is the way it is written on my sheet
anonymous
  • anonymous
the top is sqrt x though
anonymous
  • anonymous
Oh gods, that makes it so much easier.
Callisto
  • Callisto
can you write it. by clicking the button 'draw' ?
anonymous
  • anonymous
Sorry I put a 4 originally but it is integral sqrt x/sqrt(x^3) + 4 dx
anonymous
  • anonymous
\[\int\frac{\sqrt x}{\sqrt{x^3}+4}\,dx\]?
anonymous
  • anonymous
Is this it?
anonymous
  • anonymous
Yes badreferences that is it.
anonymous
  • anonymous
Alright, it should be fairly straightforward. Let \(u=\sqrt{x^3}\).
anonymous
  • anonymous
Which makes \(du=\frac{3x^2}{2\sqrt{x^3}}\,dx\)
anonymous
  • anonymous
She said if it were a test would be worth 10 marks so figure there is alot to it.
anonymous
  • anonymous
So, uh, need any further help?
Callisto
  • Callisto
I dunno if it is like that
1 Attachment
anonymous
  • anonymous
I hate doing this, but seriously, I'm not typing out the entire solution. http://www.wolframalpha.com/input/?i=integrate+sqrtx%2F%28sqrt{x^3}%2B4%29
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=integrate+sqrtx%2F%28sqrt(x^3)%2B4%29
anonymous
  • anonymous
Calistro that looks like the type of stuff she is looking for.
Callisto
  • Callisto
i think your question worths.... 2 or 3 marks here... when you say it worths 10.. i think i made mistakes
anonymous
  • anonymous
Her instructions exactly were you have to use algabraic integration to delete all square roots.
Callisto
  • Callisto
@badreferences the du=....dx you've written can actually be simplified
anonymous
  • anonymous
Sure, I believe that. I generally don't simplify things though; makes the theory much nicer at the end. :)
anonymous
  • anonymous
Yeah she is pretty particular. Wants no square roots. We tried to do it in class then stopped everyone and said have to use algebraic substituion to get rid of ALL square roots.
Callisto
  • Callisto
but it makes the answer ugly :S
anonymous
  • anonymous
This is a new teacher she is very by the book and wants only full answers used. Yes Really ugly question and answer lol
anonymous
  • anonymous
That's because the professor is not good at asking questions. Or apparently teaching. There's nothing particular about getting rid of square roots; either it's the only way to solve it, or it's a dumb habit.
anonymous
  • anonymous
New teacher, dumb habits and dumb teaching. It`s her first year
Callisto
  • Callisto
\[du = (3/2)x^{(3/2)-1}dx\]\[du = (3/2)x^{(1/2)}dx\] What I've learnt is to reduce the power by one...
anonymous
  • anonymous
Makes no difference, treating it as a power rule or a function rule.
Callisto
  • Callisto
But surely the answer looks prettier and i think you should keep the answer in the simplest form?
anonymous
  • anonymous
Annoying habit. I try to ignore it. Complexifying equations, being able to see its parts no matter what form it's in, is important in analytical mathematics.
anonymous
  • anonymous
Callistro yep that is the format to reduce by one according to my teachers method. I see what you are saying thought bad references.

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