## anonymous 5 years ago how do i integrate dx/sqrt(4-x)

1. anonymous

Have you tried u substition?

2. anonymous

substitution as in integration by parts?

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. 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

7. anonymous

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

8. anonymous

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

9. anonymous

sorry getting used to the equation typing

10. 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.

11. anonymous

why is du/dx = -1?

12. 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.

13. anonymous

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

14. anonymous

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