anonymous
  • anonymous
integrate 1/9+sqrt(2x) I'm supposed to solve it using u-substitution but im stumped
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{9+\sqrt{2x}}~dx}\)
anonymous
  • anonymous
yes
freckles
  • freckles
I dare you to try subbing the whole bottom as u

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More answers

anonymous
  • anonymous
that is du = 1/sqrt(2x) dx but what do i do about the sqrt(2x) ?
freckles
  • freckles
\[u=9+\sqrt{2x} \\ u-9=\sqrt{2x}\]
freckles
  • freckles
\[\int\limits \frac{u-9}{u} du\]
anonymous
  • anonymous
oh ok that makes alot of sense
anonymous
  • anonymous
thank you
anonymous
  • anonymous
x - 9(ln|9 + sqrt(2x)|) + C
freckles
  • freckles
just in case... \[u=9+\sqrt{2x} \\ du=\frac{1}{\sqrt{2x}} dx \\ \text{ replace } \sqrt{2x} \text{with } u-9 \\ du=\frac{1}{u-9} dx \\ (u-9) du =dx\] ok now checking your final answer...
freckles
  • freckles
why did you replace the first u with x and not 9+sqrt(2x)?
freckles
  • freckles
should have u-9ln|u|+C where u=9+sqrt(2x)
freckles
  • freckles
that first u you put x?
anonymous
  • anonymous
can't you split it up into u/u - 9/u ?
freckles
  • freckles
you can
freckles
  • freckles
but you are integrating with respect to u not one x and the other u
anonymous
  • anonymous
oh ok so the 1 would integrate back into u which i would then replace
freckles
  • freckles
right because we are integrating with respect to u because of the du part
freckles
  • freckles
\[\int\limits (1-\frac{9}{u} ) du\\ u-9 \ln|u|+C\]
freckles
  • freckles
and final step is to replace any u you see with 9+sqrt(2x)
anonymous
  • anonymous
ok thank you, I'll practice more problems like this one
freckles
  • freckles
k
freckles
  • freckles
I think there is another way without u sub to do this one would you like to see?
anonymous
  • anonymous
yes
anonymous
  • anonymous
please
freckles
  • freckles
\[\frac{1}{9+\sqrt{2x}} \cdot 1 \\=\frac{1}{9+\sqrt{2x}} \cdot \frac{9-\sqrt{2x}}{9-\sqrt{2x}} \\ =\frac{9-\sqrt{2x}}{81-2x}\] oops spoke to soon.
anonymous
  • anonymous
i actually tried doing it that way but i quickly figured out i would involve multiple substitutions which idk how to do yet
freckles
  • freckles
\[\frac{1}{9+\sqrt{2x}} =\frac{1}{\sqrt{2x}} \cdot \frac{\sqrt{2x}}{9+\sqrt{2x}} \\ =\frac{1}{\sqrt{2x}} \cdot \frac{\sqrt{2x}+9-9}{9+\sqrt{2x}} \\ =\frac{1}{\sqrt{2x}} \cdot (\frac{\sqrt{2x}+9}{9+\sqrt{2x}}-\frac{9}{9+\sqrt{2x}}) \\ =\frac{1}{\sqrt{2x}} (1-\frac{9}{9+\sqrt{2x}}) \\ =\frac{1}{\sqrt{2x}}- \frac{1}{\sqrt{2x}} \cdot \frac{9}{9+\sqrt{2x}}\] yea you still might prefer a substitution on the second one
freckles
  • freckles
so yep I spoke way too soon earlier my bad
anonymous
  • anonymous
hahaha no problem but seriously thank you for the help
freckles
  • freckles
np

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