Tae B.

asked • 07/24/14

(x^2+1)/(x^4+4)

Determine this intergral by expanding it first

Bob A.

Did a professor Really give you this integral?
It is a ridiculously hard one to be used for practice.  And expanding it 1st is not the easiest method.  Doing it all together you can combine some terms as you go and instead of 20 steps per term plus some to combine at the end - you can do it in a little over 20 steps total.
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07/24/14

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Bob A. answered • 07/24/14

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To expand you would expand it into a polynomial - if multiplication you would multiply by the distribution method (sometimes Wrongly called the foil method) - for division like here you would separate the terms of the numerator dividing each by a common denominator.
 
∫  (x2+1)   =  ∫    x2      +  ∫     1       
   (x4+4)           (x4+4)          (x4+4)
 
So now you can find the integral of each term
factor the denominators into linear irreducible quadratic terms
(note: you could do this 1st before separating terms but you said expand 1st)
 
(x4+4) = (x2 - 2x + 2)(x2 + 2x +2)
 
Integrate Term by Term
- for the 1st (x2) term   use partial fractions
 
x2 / (x2 - 2x + 2)(x2 + 2x +2)
 
 = x / 4(x2 - 2x + 2) - x / 4(x2 + 2x +2)
 
I did 20 more steps for each term  plus some to put the terms together to get to
 
1/16  [  ln(x2-2x+2)  -  ln(x2+2x+20)  -  6tan-1(1-x)  +  6tan-1(x+1)  ] + C
 
Sorry but it would take all day to type it into this foolish window that WyzAnt has with such poor character handling for math equations.
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Bob A.

It was easier to take the step by step from Wolfram Alpha
What is below is a copy and paste from Wolfram Alpha
They don't expand the terms first - but show the end result.
 
Take the integral:
integral (x^2+1)/(x^4+4) dx

For the integrand (x^2+1)/(x^4+4),
factor the denominator into linear and irreducible quadratic terms:
= integral (x^2+1)/((x^2-2 x+2) (x^2+2 x+2)) dx

For the integrand (x^2+1)/((x^2-2 x+2) (x^2+2 x+2)),
use partial fractions:
= integral ((2-x)/(8 (x^2+2 x+2))
     + (x+2)/(8 (x^2-2 x+2))) dx

Integrate the sum term by term and factor out constants:
= 1/8 integral (2-x)/(x^2+2 x+2) dx
     + 1/8 integral (x+2)/(x^2-2 x+2) dx

Rewrite the integrand (2-x)/(x^2+2 x+2)
as 3/(x^2+2 x+2) - (2 x+2)/(2 (x^2+2 x+2)):
= 1/8 integral (3/(x^2+2 x+2) - (2 x+2)/(2 (x^2+2 x+2))) dx
     + 1/8 integral (x+2)/(x^2-2 x+2) dx

Integrate the sum term by term and factor out constants:
= - 1/16 integral (2 x+2)/(x^2+2 x+2) dx
     + 3/8 integral 1/(x^2+2 x+2) dx
     + 1/8 integral (x+2)/(x^2-2 x+2) dx

For the integrand (2 x+2)/(x^2+2 x+2),
substitute u = x^2+2 x+2 and du = (2 x+2) dx:
= - 1/16 integral 1/u du + 3/8 integral 1/(x^2+2 x+2) dx
     + 1/8 integral (x+2)/(x^2-2 x+2) dx

The integral of 1/u is log(u):
= - (log(u))/16 + 3/8 integral 1/(x^2+2 x+2) dx
    +1/8 integral (x+2)/(x^2-2 x+2) dx

For the integrand 1/(x^2+2 x+2), *** See Footnote
complete the square:
= - (log(u))/16 + 3/8 integral 1/((x+1)^2+1) dx
     + 1/8 integral (x+2)/(x^2-2 x+2) dx

For the integrand 1/((x+1)^2+1),
substitute s = x+1 and ds = dx:
= - (log(u))/16 + 3/8 integral 1/(s^2+1) ds
     + 1/8 integral (x+2)/(x^2-2 x+2) dx

The integral of 1/(s^2+1) is tan^(-1)(s):
= 3/8 tan^(-1)(s) - (log(u))/16
    + 1/8 integral (x+2)/(x^2-2 x+2) dx

Rewrite the integrand (x+2)/(x^2-2 x+2)
   as (2 x-2)/(2 (x^2-2 x+2)) + 3/(x^2-2 x+2):
= 3/8 tan^(-1)(s) - (log(u))/16
    + 1/8 integral ((2 x-2)/(2 (x^2-2 x+2))
    + 3/(x^2-2 x+2)) dx

Integrate the sum term by term and factor out constants:
= 3/8 tan^(-1)(s) - (log(u))/16
   + 1/16 integral (2 x-2)/(x^2-2 x+2) dx
   + 3/8 integral 1/(x^2-2 x+2) dx

For the integrand (2 x-2)/(x^2-2 x+2),
substitute p = x^2-2 x+2 and dp = (2 x-2) dx:
= 3/8 tan^(-1)(s) - (log(u))/16
    + 1/16 integral 1/p dp
    + 3/8 integral 1/(x^2-2 x+2) dx

The integral of 1/p is log(p):
= 3/8 tan^(-1)(s)  +  (log(p))/16  -  (log(u))/16
    + 3/8 integral 1/(x^2-2 x+2) dx

For the integrand 1/(x^2-2 x+2),
complete the square:
= 3/8 tan^(-1)(s)  +  (log(p))/16  -  (log(u))/16
    + 3/8 integral 1/((x-1)^2+1) dx

For the integrand 1/((x-1)^2+1),
substitute w = x-1 and dw = dx:
= 3/8 tan^(-1)(s)  +  (log(p))/16  -  ( log(u))/16
     + 3/8 integral 1/(w^2+1) dw

The integral of 1/(w^2+1) is tan^(-1)(w):
= (log(p))/16  +  3/8 tan^(-1)(s)
      - (log(u))/16  +  3/8 tan^(-1)(w) + constant

Substitute back for w = x-1:
= (log(p))/16  +  3/8 tan^(-1)(s)
     - (log(u))/16  -  3/8 tan^(-1)(1-x) + constant

Substitute back for p = x^2-2 x+2:
= 3/8 tan^(-1)(s)  -  (log(u))/16 + 1/16 log(x^2-2 x+2)
     -  3/8 tan^(-1)(1-x) + constant

Substitute back for s = x+1:
= -(log(u))/16  +  1/16 log(x^2-2 x+2)
     -  3/8 tan^(-1)(1-x)  +  3/8 tan^(-1)(x+1)  + constant

Substitute back for u = x^2+2 x+2:
= 1/16 log(x^2-2 x+2)  -  1/16 log(x^2+2 x+2)
     3-  /8 tan^(-1)(1-x)  +  3/8 tan^(-1)(x+1)  + constant

Which is equal to:
Answer:
= 1/16 (log(x^2-2 x+2)  -  log(x^2+2 x+2)
     -  6 tan^(-1)(1-x)  +  6 tan^(-1)(x+1))  + constant
 
*** I found a little footnote in Wolfram Alpha saying that they do not use standard notation!
Log[z]     gives the natural logarithm of z (logarithm to base ).
Log[b,z]    gives the logarithm to base b.
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07/24/14

Tae B.

Thank you so much
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07/25/14

SURENDRA K. answered • 07/24/14

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(x^2+1)(x^4+4)
 
= x^6+x^4+4x^2+4
 
Take integral of above
 
= (x^7)/7. +  (x^5)/5  +(4x^3)/3+4x+c(constant)
 
 
 
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Kee B.

I still dont get how you got this
 
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07/24/14

SURENDRA K.

No, I took a different question.
Your question is division of two polynomial & not multiplication.
I take back my solution.
 
 
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07/24/14

Kee B.

ok thanks anyway
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07/24/14

SURENDRA K.

Hello Robert,
 
Kindly verify the last step, what I am getting is as follows.
 
(1/8)[ln[(x^2-2x+2)/(x^2+2x+2)]]+(1/4)*taninverse(x-1)+1/4*taninverse(x+1)+c
 
 
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07/24/14

Bob A.

I did make a mistake in the input of the result.
 
1/16 [ logx2-2x+2) - log(x2+2x+20 - 6tan-1(1-x) + 6tan-1(x+1) ] +C
 
Should Be
 
1/16 [ logx2-2x+2) - log(x2+2x+2) - 6tan-1(1-x) + 6tan-1(x+1) ] +C
 
But I think the rest is correct.
I also put it into Wolfram Alpha and it gives the same answer.  But it wouldn't be the first time I have found errors in Wolfram.
I'll go put my steps as a comment on my original answer.
Give me a bit of time to get it typed in, and please let me know if I made a mistake,
 
@ WyzAnt
I wish that copy and paste in the answers section would copy the formatting - it would speed up doing math answers TREMENDOUSLY. I hope you heard that WA.
Of course it would be even better if you took standard keyboard input instead of having to use your hokey symbol thing under the omega in the toolbar !!
 
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07/24/14

SURENDRA K.

Well,I have again gone through my solution.
What I am getting is as follows please.
 
(1/16)[ln[(x^2-2x+2)/(x^2+2x+2)]]+(3/8)*taninverse(x-1)+(3/8)*taninverse(x+1)+c
 
The only difference lies in the coefficients of taninverse terms.
 
 
 
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07/24/14

Bob A.

What you have is:
(1/16) { ln[(x^2-2x+2) / (x^2+2x+2)] }  +   (3/8)*taninverse(x-1)   +   (3/8)*taninverse(x+1)   +   c
 
Where the 1.16 is times only the LN term
If you combine the 1/16  with the 3/8's of the two arcTan terms
you get an equivalent form  since 6*1/16 = 3/8
 
(1/16)  { ln[(x^2-2x+2) / (x^2+2x+2)] + (6)*arcTan(x-1) + (6)*arcTan(x+1) }   + c
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07/24/14

SURENDRA K.

OH ---- it is too much.
I did not notice ---
1/16  is outside the bracket.
 
 
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07/25/14

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