Question

make a substitution to express the integrand as a rational function and then evaluate the integral
\[\int_{9}^{16} \frac{\sqrt(x)}{x-4} dx\]

Answer 1

isn't it already a rational function?

Answer 2

no

Answer 3

rational function is not one function over another, it is the ratio of two polynomials

Answer 4

oh ok

Answer 5

i guess you should try \(u=\sqrt{x}\)

Answer 6

ok

Answer 7

that would make things more complicated i think

Answer 8

then don't be scared because \(du=\frac{1}{2\sqrt{x}}dx\)

Answer 9

no it makes it easier

Answer 10

\[\frac {u}{u^2 -4}\]

Answer 11

since \(u=\sqrt{x}\) you get
\(2\sqrt{x}du=dx\) or
\[2u=dx\]

Answer 12

what about the dx?

Answer 13

lets go back to the start

Answer 14

i see it now...

Answer 15

\[u=\sqrt{x}\]
\[du=\frac{1}{2\sqrt{x}}dx\]
\[2\sqrt{x}du=dx\] and since \(u=\sqrt{x}\) this becomes \(2udu=dx\)

Answer 16

\frac {2u^2}{u^2 -4}

Answer 17

\[\frac {2u^2}{u^2 -4}
\]

Answer 18

so your integral is \[\int\frac{u^2du}{u^2-4}\]

Answer 19

oops i dropped the 2
you are right
don't forget to change the limits of integration

Answer 20

plug in 16 and 9 to sqrtx ?

Answer 21

yes, exactly
doesn't what work out nicely?

Answer 22

yep...thanks again

Answer 23

yw
integral should be ok from there right?
start with
\[2\int_3^4\frac{u^2du}{u^2-4}\] and divide

Answer 24

actually there is still some annoying pfd to do , but it is doable