Question

\(\LARGE \int \frac{4x+3}{\sqrt{5 + 9x - x^2 }} dx\)

is this completing the square then inverse trig?

Answer 1

here it is @Mimi_x3 :D

Answer 2

looks like complete the square then a u-sub

Answer 3

wait..let me try it first..

Answer 4

okay then :D though completing the square seems tough since itll be (9/2 -x)^2 which looks complicated

Answer 5

well, have you completed the square first

Answer 6

do you mean in this problem? hmm like i said it seemed complicated

Answer 7

i'll have \(\LARGE \sqrt{(\frac{9}{2} - x)^2 -\frac{61}{4}}\) lol

Answer 8

wolfram alpha says its..
\[\frac{101}{4}-\left(x-\frac{9}{2}\right)^{2} \]
too tired to do it..lol

Answer 9

o.O

Answer 10

well then assuming that's it...then.. this is arcsin right...

Answer 11

how do i get rid of the numerator? one of my common problems in these types

Answer 12

You did something wrong when you completed the sqaue.
\[\left(-x+\frac{9}{2}\right) ^{2} -\left(\frac{9}{2}\right) ^{2} +5\]
then..
\[\frac{101}{4}-\left(x-\frac{9}{2}\right)^{2} \]

Answer 13

Then you have to u-sub

Answer 14

i dont see the u sub lol

Answer 15

\[\large \frac{4x+3}{\sqrt{(-x+\frac{9}{2})^{2}-\frac{101}{4}}} \]

Answer 16

hmm..

Answer 17

wiat

Answer 18

breaking it up into two fractions might work

Answer 19

brb 5 min

Answer 20

\(\LARGE \int \frac{4x}{\sqrt{(-x + \frac{9}{2})^2 - \frac{101}{4}}} + \frac{3}{\sqrt{(-x + \frac{9}{2})^2 - \frac{101}{4}}}\)

Answer 21

if u = -x + \(\frac{9}{2}\) then du = -dx that does not satisfy 4x:P

Answer 22

will work for the second one i will have to think about the first

Answer 23

for the second one it will be a trig sub right?

Answer 24

its going to get into a big mess..i think that its a trig sub

Answer 25

i think the first one is a trig sub

Answer 26

lol this is complicated =))

Answer 27

no; the second one is easy..
man i can't think..too tired..i think there is a better way..

Answer 28

this looks so complicated to me :/ yet i cannot see any other way

Answer 29

hmph; i give up im going to sleep..will think of a better way later if no one answers it..

Answer 30

okay thanks for the help mimi :)

Answer 31

why are you using \[\sqrt{(-x+\frac{9}{2})^{2}-\frac{101}{4}}\] as the bottom of your integrand?

Answer 32

to make it look like an arcsin?

Answer 33

i mean the one with 1/a arcsin u/a

Answer 34

but when you complete the square you get \[\frac{101}{4}-\left(x-\frac{9}{2}\right)^{2}\]

Answer 35

\[\frac{2(2x-9) + 18 + 3}{\sqrt{5 + 9x - x^2}} = 2\frac{2x - 9}{\sqrt{5 + 9x - x^2}} + \frac{21}{\sqrt{5 + 9x - x^2}}\]
\[5 + 9x - x^2=t^2 \implies 2t \frac{dt}{dx} =9 -2x \]
Perform the t substitution in the first one and complete squares in the second one. It should work.

Answer 36

don't you have it backwards

Answer 37

what did ishaan do the last step

Answer 38

Hmm? lgb? I tried to cancel (get rid of the x term) the numerator by the substitution of the polynomial under square root.

Answer 39

i meant this 5+9x−x^2=t^2

Answer 40

Basic substitution. I didn't want to have square roots.

Answer 41

after the substitution.*

Answer 42

oh i see...well how you gonna plug in 9 - 2x:P

Answer 43

\[2\frac{2x - 9}{\sqrt{5 + 9x - x^2}} dx = 2\frac{2x - 9}{\sqrt{5 + 9x - x^2}} \cdot \frac{2t}{9-2x}\cdot dt = 2\cdot \frac{-(9-2x)}{t}\cdot \frac{2t}{9-2x}dt\]