Integration Question

How do you integrate

(sqrt(x)) / (sqrt(Ax-B)) where A and B are constants (assume positive if it makes any difference)

??

i got the answer and it involves logs but i have no idea how to get their? I've tried a few trig and hyperbolic substitutions but got nowhere

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Reply 1

alex2100x

Have you met this before?

\displaystyle \int \frac{f'(x)}{f(x)}dx=ln|f(x)|+c

Reply 2

Jammy4410

Original post by alex2100x

Have you met this before?

\displaystyle \int \frac{f'(x)}{f(x)}dx=ln|f(x)|+c

yeah, not sure how it helps though...

Reply 3

Mr M

Study Forum Helper

Original post by Jammy4410

yeah, not sure how it helps though...

Me neither.

Reply 4

TeeEm

Original post by Jammy4410

Original post by Mr M

Where did you get your answer from?!

Try differentiating

x^{\frac{3}{2}} (Ax-B)^{-{\frac{1}{2}}}

these are usually done by hyperbolic/trigonometric substitutions

(edited 9 years ago)

Reply 5

steve44

Should you write it as

{\sqrt{x} \over \sqrt{Ax-B}} = {x \over \sqrt{Ax^2 - Bx}} = {C (2Ax - B) + D \over \sqrt{Ax^2-Bx}} = C {2Ax - B \over \sqrt{Ax^2 - Bx}} + D {1 \over \sqrt{A} \sqrt{(x - B/2A)^2 - B^2 / 4A^2}}

first?

(edited 9 years ago)

Reply 6

steve44

The second integral will give an inverse cosh which can be written in a form involving a log.

Reply 7

Mr M

Study Forum Helper

Original post by TeeEm

these are usually done by hyperbolic/trigonometric substitutions

I made a silly mistake.

Reply 8

Jammy4410

Original post by steve44

Should you write it as

{\sqrt{x} \over \sqrt{Ax-B}} = {x \over \sqrt{Ax^2 - Bx}} = {C (2Ax - B) + D \over \sqrt{Ax^2-Bx}} = C {2Ax - B \over \sqrt{Ax^2 - Bx}} + D {1 \over \sqrt{A} \sqrt{(x - B/2A)^2 - B^2 / 4A^2}}

first?

I read somewhere I should multiply top and bottom by root x, so that started me off but I don't get the 2nd equality?
I got something close to the answer after completing the square then using the substitution:
x - (B)/(2A) = (B)/(2A)coshu

(edited 9 years ago)

Reply 9

Jammy4410

I think it actually just checking working again

Reply 10

steve44

Original post by Jammy4410

The second equality is true when C=1/2A and D=B/2A.

It is easy to do the first integral with the substitution y=AX^2 - Bx. The second integral from using the substitution x-B/2A = B/2A cosh u.

That is the way I would do it.

(edited 9 years ago)

Reply 11

steve44

We could do it your way and we would end up evaluating the sinh of an inverse cosh?

(edited 9 years ago)

Reply 12

Mr M

Study Forum Helper

Original post by steve44

We could do it your way and we would end up evaluating the sinh of an inverse cosh?

That's usually not difficult.

http://en.wikipedia.org/wiki/Inverse_hyperbolic_function#Composition_of_hyperbolic_and_inverse_hyperbolic_functions

Reply 13

steve44

Original post by Mr M

That's usually not difficult.

http://en.wikipedia.org/wiki/Inverse_hyperbolic_function#Composition_of_hyperbolic_and_inverse_hyperbolic_functions

I just realised that.

Reply 14

the bear

you can apply various substitutions and end up with

K ∫ sec³ θ dθ

which can be integrated by parts:

http://upload.wikimedia.org/math/5/0/6/5066f81897ec518737dbf3bda5c584ff.png

you need to integrate sec θ ....

http://math2.org/math/integrals/more/sec.htm

:borat:

Reply 15

Jammy4410

Original post by steve44

I just realised that.

So nearly got the answer, I did the working my way since your method is probably a step too clever for me to see if I was given another similar integral.

so I got:

x / ( root[A] x root [( x-(B/2A) )^2 - (B^2)/(4A^2)] )

then i used the substitution x - B/2A = (B/2A) x Coshu

this simplified the integral to:

x/ root[A] = B/(2A)^1.5 x (coshu + B)

which integrated to

B/(2A)^1.5 x (Sinhu +u) +const

Reply 16

Jammy4410

then i subbed in for sinhu = root[ (2AX-B)^2/(B^2) -1]
and also u =arcosh [(2AX-B)/(B)]

Reply 17

Jammy4410

the first term B/2A^1.5 x sinhu simplified to root[x] X root[Ax-B] all over A ---- which was right

Reply 18

Jammy4410

Original post by the bear

worked perfectly and confirmed the answer I had was right. Don't understand why my method didn't work to the end...

Reply 19

Jammy4410

just finally checked that doing it using the substitution involving cosh will get to the same correct answer but being able to simplify to the answer is almost impossible to spot unless you work back from the answer