Question

im missing a bit of algeba here. . .

integrate 3 / (2-sqrt x) if i use u = sqrt x then du is
1/(2 sqrt x) when wolfram does the sub it writes the integral as
6 integral u / (2-u) i understand everything except the u in the numerator. how is that u in the numerator getting there

Answer 1

\[\int \frac{3}{2-\sqrt(x)}dx\]

Answer 2

this is it right?

Answer 3

if we claim that u = sqrt(x), then du = 1/2sqrt(x) dx; and since sqrt(x) IS u, we can rewrite this as, 1/2u ... if that helps

Answer 4

so lets apply this;

du = 1/2u dx

2u du = dx

Answer 5

now replace what gets replaced

Answer 6

the u is in the denominator because we let u=sqrtx which is being substituted in the denominator

Answer 7

int 3 / 2-sqrt(x) dx

int 3 * 2u / 2-u du

Answer 8

\[\int 6\frac{u}{2-u}du\]

Answer 9

all were missing now is a negtive and we are good to go

Answer 10

3 / (2-sqrt x) dx, u=x^(1/2) so du=(1/2)x^(-1/2)dx so dx=2x^(1/2)du and if you substitute this in here 3 / (2-sqrt x) dx, you get (3/(2-u))(2x^(1/2))du, since u=x^(1/2) then (3/(2-u))(2x^(1/2))du becomes (3/(2-u))(2u)du which is
(6u/2-u)du..this should answer your question..please reply back if you understand this thanks

Answer 11

that could have been formated better ...

Answer 12

from above ....

3 / (2-sqrt x) dx

u=x^(1/2) ; du=(1/2)x^(-1/2)dx

dx=2x^(1/2)du

if you substitute this in here: 3 / (2-sqrt x) dx, you get:

(3/(2-u))(2x^(1/2))du,

since u=x^(1/2) then (3/(2-u))(2x^(1/2))du becomes

(3/(2-u))(2u)du

which is (6u/2-u)du..

this should answer your question..please reply back if you understand this thanks....

Answer 13

that helps very much. Thank you both. Sorry I dropped out yesterday. I was trying the site for a tablet and it was not working so well. Thank you again.