Partial Fraction Decomposition??How do I separate this by partial fraction decomposition? ∫ 1 u ( u 2 + 1 ) d uI've used the normal technique and got to: 1 = A ( u 2 + 1 ) + B ( u ) 1 = A ( u 2 + 1 ) + B ( u )and A = 1 if u = 0 BUT how do I find B now? because I can't make u 2 + 1 equal to 0.

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Partial Fraction Decomposition??How do I separate this by partial fraction decomposition?  ∫      1          u      (              u        2            +      1      )        d  uI&#039;ve used the normal technique and got to:  1  =  A  (      u    2    +  1  )  +  B  (  u  )  1  =  A  (      u    2    +  1  )  +  B  (  u  )and   A  =  1 if   u  =  0 BUT how do I find   B now? because I can&#039;t make       u    2    +  1 equal to 0.

lavintisqpsnb · Accepted Answer

Patial fractions work when you want to split      1          P      (      x      )      Q      (      x      )      into      A          P      (      X      )        +      B          Q      (      X      )      where   P and   Q are polynomials.You don&#039;t have that. You have just one polynomial in the denominator.Hint:To calculate the integral, review the derivatives of the basic trigonometric functions and their inverses.AFTER YOUR EDIT:You made a mistake where if the denominator polynomial has degree   n, then the numerator needs to have degree   n  −  1, so in your case, you should set      1          u      (              u        2            +      1      )        =      A    u    +            B      u      +      C                      u        2            +      1      Now it should be easy to get the solution.

uto2rimxrs50 · Accepted Answer

Solution without partial Decomposition::  I  =  ∫      1          u      (              u        2            +      1      )        d  u  =  ∫      1                  u        3            (      1      +              u                  −          2                    )        d  uNow Put   1  +      u          −      2        =  t    , Then       −          2              u        3              d    u    =    d    t    ⇒          1              u        3              d    u    =    −          1      2        d    t  So  I  =  −      1    2    ∫      1    t    d  t  =  −      1    2    ln  ⁡  t  +      C    =  −      1    2    ln  ⁡      |                  1        +                  u          2                            u        2              |    +      C    =      1    2    ln  ⁡      |                  u        2                    1        +                  u          2                      |    +      C

Partial Fraction Decomposition?? How do I separate this by partial fraction decomposition? &

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