Question

quick question: \[\int\limits\frac{\sqrt{x}}{ \sqrt{x}-3}dx\]little help? :3

Answer 1

am I supposed to u sub?
if so, what?

Answer 2

\[\int\limits \frac{ { \sqrt{x}-3+3 } }{ \sqrt{x}-3 }dx\]
break in two and then ?
\[put~\sqrt{x}-3=t,\sqrt{x}=t+3,x=t^2+6t+9,dx=(2 t+6)dt\]

Answer 3

ummm I only understood the -3 + 3
@surjithayer could you explain what you did next?

Answer 4

\[\int\limits \frac{ \sqrt{x} -3}{ \sqrt{x}-3}dx+3 \int\limits \frac{ dx }{ \sqrt{x}-3 }\]

Answer 5

ok
so it becomes 1 + 3 int 1/(root(x) - 3) dx
what do I do with the int (1/(root(x)-3))dx ?

Answer 6

wait sorry int 1 dx = x

Answer 7

\[\int\limits 1 ~dx+3 \int\limits \frac{( 2t+6)dt }{ t }+c\]

Answer 8

how did you get from 3 int 1/(root(x) - 3) to that? :/

Answer 9

\[\int\limits \frac{ (\sqrt{x}-3)+3 }{ (\sqrt{x}-3 }dx=\int\limits \frac{ \sqrt{x}-3 }{ \sqrt{x}-3 }dx+\int\limits \frac{ 3 }{ \sqrt{x} -3}dx\]

Answer 10

yes but I meant
3 int 1/(root(x) - 3) to that 3 int (2t+6)dt / t

Answer 11

i have explained earlier
put \[\sqrt{x}-3=t\]
\[\sqrt{x}=t+3\]
squaring both sides
\[x=t^2+6t+9\]

Answer 12

*thinks* hmmm ok I think I got it
thank you @surjithayer

Answer 13

if there is any doubt ,you can ask me.

Answer 14

@surjithayer

I got x + 6(root(x)-3) + 18 ln|root(x)-3| + constant
but wolfram alpha says: http://www.wolframalpha.com/input/?i=integral+of+%28root%28x%29%29%2F%28root%28x%29-3%29+dx

I did
\[\int\limits 1 dx + 3\int\limits \frac{ 2t+6 }{ t }dt\]\[x + 3 \int\limits 2~du + 18 \int\limits \frac{ du }{ u }\]\[x + 6u + 18 \ln|u| + C\]\[x + 6(\sqrt{x}-3) + 18\ln|\sqrt{x}-3|+C\]

Answer 15

?!
ohhhhhh wait...
if I distribute it, it'll just become part of the constant?

Answer 16

both are correct
because \[6\left( \sqrt{x}-3 \right)+c=6\sqrt{x}-18+c=6\sqrt{x}+C\]

Answer 17

got it, thank you so much! :D

Answer 18

yw