anonymous
  • anonymous
∫√(1+(e powered by 2)dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[1+e^2\]
anonymous
  • anonymous
?
anonymous
  • anonymous
e to the x

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anonymous
  • anonymous
you're missing a square root
anonymous
  • anonymous
cambell
anonymous
  • anonymous
the contents are enclosed in a root sing
campbell_st
  • campbell_st
then its even easier as \[\sqrt{1 + e^2}\] is a constant \[\int\limits \sqrt{1 + e^2} dx = x \sqrt{1 + e^2} + c\]
anonymous
  • anonymous
help me guys
anonymous
  • anonymous
how do i then work it out
anonymous
  • anonymous
^ what cambell said. you have a constant unless that last part is e^2x then you have something different
anonymous
  • anonymous
its actually e raised to x
anonymous
  • anonymous
not to 2
campbell_st
  • campbell_st
well \[\sqrt{1 + e^2} = 2.896\] its just a number.... you can plot it on the number line the the problem is just like \[\int\limits 5 dx = 5x + c\]
anonymous
  • anonymous
\[\int\limits_{}^{}\sqrt{1+e ^{x}}\]
campbell_st
  • campbell_st
lol... now you have the question \[\int\limits \sqrt{1 + e^{2x}} dx\]
anonymous
  • anonymous
only x not 2x
anonymous
  • anonymous
dnt leave yet campbell i could really use your help with a couple qsns
anonymous
  • anonymous
is it goin well
campbell_st
  • campbell_st
well then its integration by substitution let u = e^x du/dx = e^x du = e^xdx or dx = du/u \[\int\limits \sqrt{(1 + u)}/u du\] looks like a double substitution... good luck
anonymous
  • anonymous
what then would be the final answer ?
Zarkon
  • Zarkon
\[u=1+e^x\]

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