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Have you tried u substition?

substitution as in integration by parts?

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

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

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

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

sorry getting used to the equation typing

why is du/dx = -1?

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

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