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Mathematics 14 Online
OpenStudy (anonymous):

if the substitution u=sqrt(x+1) is used, then dx/(xsqrt(x+1)) on the interval [0,3] is equivalent to...

OpenStudy (anonymous):

1/2sqrt(x+1)

ganeshie8 (ganeshie8):

\[\rm u = \sqrt{x+1} \implies du = \frac{1}{2\sqrt{x+1} }dx \implies \bbox[border:2px solid green]{2du = \frac{1}{\sqrt{x+1}}dx}\]

ganeshie8 (ganeshie8):

so you can replace dx/sqrt(x+1) wid 2du what about x ?

OpenStudy (anonymous):

idk what to do with the remaining x

ganeshie8 (ganeshie8):

the goal is to replace everything with \(\rm u\)'s

OpenStudy (anonymous):

u^2-1?

ganeshie8 (ganeshie8):

Yes!

ganeshie8 (ganeshie8):

plug them in

OpenStudy (anonymous):

so then it is \[2\int\limits u ^{2}-1 du\]

ganeshie8 (ganeshie8):

x is in bottom right ?

OpenStudy (anonymous):

actually \[\int\limits \frac{ 2du }{ u ^{2}}\]

OpenStudy (anonymous):

-1

OpenStudy (anonymous):

on interval 2,1

OpenStudy (anonymous):

thank you!

OpenStudy (anonymous):

that's all i need to do for the answer

ganeshie8 (ganeshie8):

interval doesnt look correct

ganeshie8 (ganeshie8):

how did u get 2,1 ?

OpenStudy (anonymous):

subbed 3 and 0 into sqrt(x+1)

ganeshie8 (ganeshie8):

Oh right !

OpenStudy (anonymous):

:)

ganeshie8 (ganeshie8):

\[\rm \int\limits_1^2 \dfrac{2du}{u^2-1}\]

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