Question

Evaluate the integral:

Answer 1

\[\int\limits_{}^{}\frac{ 1 }{ x^2\sqrt{12x+1} }dx\]

Answer 2

I got to the point where I have to use partial fractions.

Answer 3

\[24\int\limits_{}^{}\frac{ 1 }{(s^2-1)^2 }ds\]

Answer 4

I am stuck here.

Answer 5

Lot's of factoring.

(s+1)(s-1) repeated.

Answer 6

My work:

\[\int\limits\limits_{}^{}\frac{ 1 }{ x^2\sqrt{12x+1} }dx\]

u=12x+1

x=(u-1)/12

1/12 du=dx

\[\int\limits_{}^{}\ \frac{ 1 }{ (\frac{ u-1 }{ 12 })^2\sqrt{u}}du\]


\[12\int\limits_{}^{}\frac{ 1 }{ (u-1)^2\sqrt{u} }du\]

s=√ (u)

s^2=u

ds=1/2√ u) du

2√ (u)ds=du

\[24\int\limits_{}^{}\frac{ 1 }{ (s^2-1)^2 }ds\]

Answer 7

I am stuck here.

Answer 8

I feel that I could integrate that easily but something stupid is preventing me from doing so.

Answer 9

I am pretty sure I have to use partial fractions but I don't know how...

Answer 10

(s^2-1)^2 = (s+1)^2 (s-1)^2
A/(s+1) +B/(s+1)^2 + C/(s-1)+ D/(s-1)^2

Answer 11

How did you break it up? @hartnn

Answer 12

generally,
(ax+b)^2 in the denominator is break up like
A/(ax+b) +B/(ax+b)^2

Answer 13

Sorry. I JUST learnt this :P .

Answer 14

the numerator should be 1 degree less than the denom for partial fractions.
so for 1/(ax+b) it'll be A/(ax+b)
and
(ax+b)^2 in the denominator is break up like
A/(ax+b) +B/(ax+b)^2

Answer 15

Allright, How did you get this though?

A/(s+1) +B/(s+1)^2 + C/(s-1)+ D/(s-1)^2

Answer 16

for 1/(s+1)^2
A/(s+1) +B/(s+1)^2
got this ?

same thing for 1/(s-1)^2

Answer 17

A/(s+1) +B/(s+1)^2

Explain that? Sorry, I am simply not understanding the partial fraction decomposition :P .

Answer 18

for 1/(x)
you would have written A/x right ?

Answer 19

Yeah, if I didn't know that was 1.

Answer 20

for 1/(x+1) ?

Answer 21

A/(x+1) ?

Answer 22

No wait.

Answer 23

B/(x+1)

Answer 24

Would have a different coefficient.

Answer 25

ok, whatever
now we go for ^2 in denominator.
how would u write 1/x^2 ?

Answer 26

(Ax+B)/x^2
or
A/x +B/x^2

Answer 27

similarly for 1/x^3
(Ax^2+Bx+C)/x^3
or
A/x +B/x^2+C/x^3

Answer 28

see the pattern ?

Answer 29

I know that by memorizing but why?

Answer 30

But yep I see that :) .

Answer 31

the numerator should be upto 1 degree less than the denom for partial fractions.

Answer 32

Yep, that makes sense.

Answer 33

k, so would u able to find A,B,C,D now ?

Answer 34

Ahh, now I see how you split it up.

Answer 35

Umm.... Maybe, let me do it out.

Answer 36

okk..

Answer 37

@hartnn : I expanded it out and everything but surely, there must be an easier way to solve for A,B,C,D ?

Answer 38

to find B put s=-1
to find D, put s=+1

Answer 39

Because I have a MASSIVE expansion.

Answer 40

Oh right!

Answer 41

to find A,C put any other values of s
say,s=0
s=2

Answer 42

But if I put B=-1 I get 1=0 which is not true.

Answer 43

put B =-1 ?

Answer 44

\[\frac{ 1 }{ (s+1)^2(s-1)^2 }=\frac{ A }{ (s+1) }+\frac{ B }{ (s+1)^2 }+\frac{ C }{ (s-1) }+\frac{ D }{ (s-1)^2 }\]

\[1=A(s+1)^2(s-1)^3+B(s+1)(s-1)^3+C(s+1)^3(s-1)^2+D(s+1)^3(s-1)\]

Answer 45

so if I put s=-1 every term becomes 0.

Answer 46

or s=1 for that matter.

Answer 47

that expansion is incorrect.

Answer 48

:( .

Answer 49

\(1=A(s+1)(s-1)^2+B(s-1)^2+C(s+1)^2(s-1)+D(s+1)^2\)

Answer 50

Let me try again...

Answer 51

So I would go

A(s+1)^2(s−1)^3 right?

Answer 52

Because I am multiplying by every other term in the denominator on the RHS.

Answer 53

no.... you rae not doing that..

Answer 54

\[\frac{1}{x^2\sqrt{12x+1}}=\frac{\sqrt{12 x+1}}{x^2}-\frac{12 \sqrt{12 x+1}}{x}+\frac{144}{\sqrt{12 x+1}} \]

Answer 55

Sigh... I am being stupid.

Answer 56

so you are not multiplying entire bc (here,(s-1)^2(s+1)^2)
just the uncommon terms.