Integration Problem

  • Thread starter sara_87
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  • #1
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Homework Statement



How do i integrate:

[tex]\int\frac{1}{(x^2+1)^2}[/tex]

?


Homework Equations



The answer is:

[tex]\frac{1}{4}(\frac{-2x}{x^2-1}-log(x-1)+log(x+1))[/tex]

The Attempt at a Solution



I dont know which method to use.
 

Answers and Replies

  • #2
Dick
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The answer you posted is correct if the integrand is 1/(1-x^2)^2. Is that what you meant? If so, try partial fractions.
 
  • #3
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Im sorry, the question is correct but the answer is:

1/2[x/(x^2+1)+arctan(x)]

here partial fractions wont work because i will still have a term with (x^2+1)^2 in the denominator
 
  • #4
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Have you tried a trig substitution? tan (theta) = x would be a good one to start with.
 
  • #5
Dick
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Im sorry, the question is correct but the answer is:

1/2[x/(x^2+1)+arctan(x)]

here partial fractions wont work because i will still have a term with (x^2+1)^2 in the denominator

Ok, then it's a trig substitution. Like x=tan(t).
 
  • #6
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how did you know what substitution to make??
 
  • #7
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Draw a right triangle and label the sides and hypotenuse according to the expression in your integral. Since you have x^2 + 1, that corresponds to the hypotenuse. Label the side opposite your acute angle as x, label the base as 1, and label the hypotenuse as sqrt(x^2 + 1). So tan(theta) = x/1, and sec^2(theta)d(theta) = dx. Use these two equations to completely convert you integral from expressions in x and dx to ones in theta and d(theta).
 
  • #8
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If you're asking how did we know to make a trig substitution, they are good bets when you have factors involving the sum or difference of squares, such as sqrt(x^2 + a^2), sqrt(a^2 - x^2), or sqrt(x^2 - a^2).
 
  • #9
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what if we have (5+x^2)^2 in the denominator?
 
  • #10
35,139
6,892


Label the opposite side as x and the base as sqrt(x), so tan(theta) = x/sqrt(5)
 
  • #11
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thank you
 

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