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if the substitution u=sqrt(x+1) is used, then dx/(xsqrt(x+1)) on the interval [0,3] is equivalent to...
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1/2sqrt(x+1)
\[\rm u = \sqrt{x+1} \implies du = \frac{1}{2\sqrt{x+1} }dx \implies \bbox[border:2px solid green]{2du = \frac{1}{\sqrt{x+1}}dx}\]
so you can replace dx/sqrt(x+1) wid 2du what about x ?
idk what to do with the remaining x
the goal is to replace everything with \(\rm u\)'s
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u^2-1?
Yes!
plug them in
so then it is \[2\int\limits u ^{2}-1 du\]
x is in bottom right ?
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actually \[\int\limits \frac{ 2du }{ u ^{2}}\]
-1
on interval 2,1
thank you!
that's all i need to do for the answer
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interval doesnt look correct
how did u get 2,1 ?
subbed 3 and 0 into sqrt(x+1)
Oh right !
:)
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\[\rm \int\limits_1^2 \dfrac{2du}{u^2-1}\]
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