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OpenStudy (anonymous):

how do i integrate dx/sqrt(4-x)

OpenStudy (anonymous):

Have you tried u substition?

OpenStudy (anonymous):

substitution as in integration by parts?

OpenStudy (anonymous):

for your question, do you mean $\int\limits_{}^{}dx/\sqrt{4-x}$

OpenStudy (anonymous):

yes thats exactly it. i am studying improper integrals so i need to find the regular integral first

OpenStudy (anonymous):

Think of the square root on the bottom not as a square root, but as $(4-x)^{1/2}$

OpenStudy (anonymous):

You will, however, have to use a u substitution, which is different the integration by parts. Integration by parts is only necessary in situations with multiplication or division of things you cannot simplify

OpenStudy (anonymous):

More explicitly, think of it as $\int\limits_{}^{}(4-x)^{-1/2}dx$

OpenStudy (anonymous):

okay, so then when you say substitute you mean, take (4-x) and substitute u. and then perform $1/n+1\times u ^ {n+1}$

OpenStudy (anonymous):

sorry getting used to the equation typing

OpenStudy (anonymous):

essentially yes. $u = (4-x)$, and $du/dx = -1$. Substitute in u and du, integrate the function I wrote above, and replace (4-x) when you're done.

OpenStudy (anonymous):

why is du/dx = -1?

OpenStudy (anonymous):

when doing u substitutions, you substitute u, then take the derivative of what you substituted, in this case (4-x), for x, which = -1 in this case.

OpenStudy (anonymous):

oh i see. so what i have then would be $-1^{1/2}/{-1/2}$

OpenStudy (anonymous):

and that becomes -(u^1/2)/1 * 2/1 which is -2sqrt u. and that becomes -2sqrt(4-x) is that right?

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