Question

okay there comes another integral that I am going to practice.

Answer 1

\[\large \int\limits_{ }^{ } \frac{~~dx}{\sqrt{6x-x^2}}\]

Answer 2

I want to complete the square under the radical.

Answer 3

\[\large \int\limits_{ }^{ } \frac{1}{\sqrt{6x-x^2}}dx\]\[\large \int\limits_{ }^{ } \frac{1}{\sqrt{x^{\color{white}{3}}}\times~\sqrt{6-x}}dx\]\[\frac{A}{\sqrt{x}}+\frac{B}{\sqrt{x-6}}\]
\[A \sqrt{x-6^{\color{white}{3}}}~+~B \sqrt{x}=1\]

B=sqrt{6}/6
A+Bsqrt(7)=1 >> A + (sqrt{6}/6) sqrt{7}=1 >> A=1-sqrt{42}/6

Answer 4

\[\large \int\limits_{ }^{ } \frac{1-\frac{\sqrt{42}}{6}}{\sqrt{x}}+\frac{\frac{\sqrt{6}}{6}}{\sqrt{x-6}}~dx\]

Answer 5

perhaps I am doing something silly, but I feel that I am weak in math everytime I use u-sub

Answer 6

\[\large \int\limits_{ }^{ } \frac{\frac{6-\sqrt{42}}{6}}{\sqrt{x}}+\frac{\frac{\sqrt{6}}{6}}{\sqrt{x-6}}~dx\]\[(\frac{6-\sqrt{42}}{6})\large \int\limits_{ }^{ } \frac{1}{\sqrt{x}}dx +\frac{\sqrt{6}}{6}\int\limits_{ }^{ } \frac{1}{\sqrt{x-6}}~dx\]

Answer 7

the factors in the denominator are not polynomials
so don't even try partial fractions :)

Answer 8

uhhh... No. Seriously, complete the square and think about trig substitutions.

Answer 9

\[\frac{\sqrt{x}(6-\sqrt{42})}{3}+\frac{\sqrt{6}}{6}\int\limits_{ }^{ } \frac{1}{\sqrt{x-6}}~dx\]

Answer 10

I already tried partial fractions... perhaps there are mistakes, but is it impossible?

Answer 11

You should not have tried partial fractions. Complete the Square in the radical.

Answer 12

\[\frac{\sqrt{6}}{6}\int\limits_{ }^{ } \frac{1}{\sqrt{x-6}}~dx\]\[\frac{\sqrt{x}(6-\sqrt{42})}{3}+\frac{(\sqrt{x-6})\sqrt{6}}{3}+C\]

Answer 13

now I will see what I get when completing the square.

Answer 14

bring it in the form of
\(\Large \int \dfrac{1}{\sqrt{b^2- (x-a^2)}}dx\)

Answer 15

I clicked somewhere and..-:(

Answer 16

\[\int\limits_{ }^{ } \frac{1}{-x^2+6x-9+9}dx\]\[\int\limits_{ }^{ } \frac{1}{-(x-3)^2+9}dx\]

SO we would be having,\[\int\limits_{ }^{ } \frac{1}{\sqrt{-(x-3)^2+9}}dx\]

Answer 17

I can guess.
\[u=x-3,~~~du=dx\]\[\int\limits_{ }^{ } \frac{1}{\sqrt{-u^2+9}}du\]

Answer 18

think of a trig substitution

Answer 19

I am guessing sin^(-1)(u/3) ?

Answer 20

+C

Answer 21

I would think of completing the square as well

Answer 22

\[\sin^{-1}(\frac{x-3}{3})+C\]

Answer 23

thats correct :)

Answer 24

surjither is putting something up I can imagine ...

Answer 25

alternate approach i guess

Answer 26

heheh good to know all the other methods :)

Answer 27

i am writing further hartn has suggested,otherwise you are also correct.
\[\int\limits \frac{ dx }{ \sqrt{a^2-(x-b)^2} }\]
put \[x-b=a \sin \theta \]

Answer 28

no always, x->infty :P

Answer 29

you could have also tried
x-3 = 3 sin u
you'd get an equivalent answer :)

Answer 30

not always*

Answer 31

yeah, retroactively I guess you can think of this substitution.... xD

Answer 32

hmm knowing other methods gets you quickly to the answer. in exams you wouldn't spend a lot of time doing one problem

Answer 33

yeah, but I meant that trying all the methods on a single problem....

Answer 34

I mean on such a stupid day, when nothing works, like Christmas, we have time for this.

Answer 35

that's not what i meant of course

Answer 36

lol

Answer 37

alright....