Question

evaluate the integral (dx/(2root(x-4)+x))

Answer 1

\[\int\limits_{ }^{ }\left(\begin{matrix}dx \\ 2\sqrt{x-4}+x\end{matrix}\right)\]

Answer 2

\[I=\int\limits \frac{ dx }{ 2\sqrt{x-4}+x}\]
rationalise the denominator

Answer 3

you got it.Then show me your work.

Answer 4

\[\frac{ 1 }{ 2 }\int\limits_{ }^{}\frac{ \sqrt{x-4}-x }{ x^2-x-4 } dx\]

Answer 5

there is a - sign i missed

Answer 6

-x^2+x is what it should be on the bottom

Answer 7

no .it is going wrong way.
put \[\sqrt{x-4}=t,x-4=t^2,x=t^2+4,dx=2t~dt\]

Answer 8

\[I=\int\limits \frac{ 2t~dt }{ 2t+t^2+4 }=\int\limits \frac{ 2t+2-2 }{t^2+2t+4 }dt\]
can you go further?

Answer 9

how do you get 2t on the denominator?

Answer 10

\[2\sqrt{x-4}=2t\]

Answer 11

ok, I had pulled out the 2 incorectly

Answer 12

int (2t+2)/(t+2)^2 -2/(t+2)^2 dt

Answer 13

\[=\int\limits \frac{ 2t+2 }{t^2+2t+4 }dt-2 \int\limits \frac{ dt }{(t^2+2t+1)+3 }\]

Answer 14

\[=\ln \left| t^2+2t+4 \right|+?\]

Answer 15

is there a substitution im not seeing? i dont know where you get that

Answer 16

\[\int\limits \frac{ dx }{ x^2+a^2 }=\frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a }\]

Answer 17

the ln(*) part

Answer 18

\[\int\limits \frac{ f' \left( x \right) }{ f \left( x \right) }dx=\ln \left| f \left( x \right) \right|\]

Answer 19

\[\frac{ d }{ dt }\left( t^2+2t+4 \right)=\frac{ d }{ dt }(t^2) +\frac{ d }{ dt }(2t)+\frac{ d }{dt }( 4)=2t+2+0\]

Answer 20

\[\int\limits \frac{ dt }{ \left( t+1 \right)^2+\left( \sqrt{3} \right)^2 }\]

Answer 21

calculate it . formula is given above.

Answer 22

now you got it or have some quiry.