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Worksheet: Integration by Partial Fractions with Repeated Linear Factors

Q1:

Consider the function 𝑓 ( π‘₯ ) = π‘₯ ( π‘₯ βˆ’ 𝑒 ) 2 which is defined on ℝ βˆ’ { 𝑒 } .

Find an antiderivative 𝐹 of 𝑓 such that 𝐹 ( 0 ) = 2 . What is 𝐹 ( 2 𝑒 ) ?

  • A 𝐹 ( π‘₯ ) = ( | π‘₯ βˆ’ 𝑒 | ) + 𝑒 𝑒 βˆ’ π‘₯ l n , 𝐹 ( 2 𝑒 ) = 2 .
  • B 𝐹 ( π‘₯ ) = ( | π‘₯ βˆ’ 𝑒 | ) βˆ’ 𝑒 𝑒 βˆ’ π‘₯ l n , 𝐹 ( 2 𝑒 ) = 0 .
  • C 𝐹 ( π‘₯ ) = ( | π‘₯ βˆ’ 𝑒 | ) βˆ’ 𝑒 𝑒 βˆ’ π‘₯ l n , 𝐹 ( 2 𝑒 ) = 2 .
  • D 𝐹 ( π‘₯ ) = ( | π‘₯ βˆ’ 𝑒 | ) + 𝑒 𝑒 βˆ’ π‘₯ l n , 𝐹 ( 2 𝑒 ) = 0 .
  • E 𝐹 ( π‘₯ ) = ( | π‘₯ βˆ’ 𝑒 | ) βˆ’ π‘₯ 𝑒 βˆ’ π‘₯ l n , 𝐹 ( 2 𝑒 ) = βˆ’ 1 .

Is it possible to find an antiderivative 𝐺 that satisfies 𝐺 ( 0 ) = 2 , where 𝐺 ( 2 𝑒 ) = 1 ? If so, what is 𝐺 ( π‘₯ ) ?

  • AYes, 𝐺 ( π‘₯ ) =  ( 𝑒 βˆ’ π‘₯ ) + 𝑒 𝑒 βˆ’ π‘₯ π‘₯ < 𝑒 , ( π‘₯ βˆ’ 𝑒 ) + 𝑒 𝑒 βˆ’ π‘₯ + 1 π‘₯ > 𝑒 . l n i f l n i f
  • Bno

What you have found above seems to violate the result that says that any two antiderivatives must differ by a constant function, because 𝐺 βˆ’ 𝐹 is not a constant function. Why is there no contradiction?

  • Abecause that result only holds sometimes; sometimes it fails
  • Bbecause neither 𝐹 nor 𝐺 is differentiable; that result only applies to differentiable functions
  • Cbecause that result assumes that the domain is an interval
  • Dbecause that result requires additional conditions to the antiderivatives
  • Ebecause 𝐺 βˆ’ 𝐹 is constant; it has the value 0 on ( βˆ’ ∞ , 𝑒 ) and the value 1 on ( 𝑒 , ∞ )

Q2:

Find 𝐹 so that 𝐹 β€² ( π‘₯ ) = π‘₯ ( π‘₯ βˆ’ 𝑒 ) 2 and 𝐹 ( 0 ) = 2 .

  • A 𝐹 ( π‘₯ ) = | ( π‘₯ βˆ’ 𝑒 ) | + 𝑒 𝑒 βˆ’ π‘₯ l n
  • B 𝐹 ( π‘₯ ) = | ( 𝑒 βˆ’ π‘₯ ) | βˆ’ π‘₯ 𝑒 βˆ’ π‘₯ l n
  • C 𝐹 ( π‘₯ ) = | ( π‘₯ βˆ’ 𝑒 ) | βˆ’ π‘₯ 𝑒 βˆ’ π‘₯ l n
  • D 𝐹 ( π‘₯ ) = | ( 𝑒 βˆ’ π‘₯ ) | + 𝑒 𝑒 βˆ’ π‘₯ l n
  • E 𝐹 ( π‘₯ ) = | ( 𝑒 βˆ’ π‘₯ ) | + 𝑒 π‘₯ βˆ’ 𝑒 l n

Q3:

Use partial fractions to evaluate ο„Έ 6 π‘₯ + 7 ( π‘₯ + 2 ) π‘₯ 2 d .

  • A 6 | π‘₯ + 2 | + 5 ( π‘₯ + 2 ) + 𝐾 l n βˆ’ 3
  • B 6 | π‘₯ + 2 | + 5 | π‘₯ + 2 | + 𝐾 l n l n
  • C 6 | π‘₯ + 2 | βˆ’ 5 ( π‘₯ + 2 ) + 𝐾 l n βˆ’ 1
  • D 6 | π‘₯ + 2 | + 5 ( π‘₯ + 2 ) + 𝐾 l n βˆ’ 1
  • E 3 | π‘₯ + 2 | βˆ’ 4 ( π‘₯ + 2 ) + 𝐾 l n βˆ’ 1

Q4:

Use partial fractions to find an analytic expression for the integral

  • A 2 ( π‘₯ ) + ( π‘₯ βˆ’ 2 ) βˆ’ 1 π‘₯ βˆ’ 2 βˆ’ 1 l n l n
  • B 2 ( π‘₯ ) + ( 2 βˆ’ π‘₯ ) βˆ’ 1 π‘₯ βˆ’ 2 l n l n
  • C 2 ( π‘₯ ) + ( 2 βˆ’ π‘₯ ) + 1 π‘₯ βˆ’ 2 βˆ’ 1 l n l n
  • D 2 ( π‘₯ ) + ( 2 βˆ’ π‘₯ ) βˆ’ 1 π‘₯ βˆ’ 2 βˆ’ 1 l n l n
  • E 2 ( π‘₯ ) + ( π‘₯ βˆ’ 2 ) + 1 π‘₯ βˆ’ 2 βˆ’ 1 l n l n

Q5:

Use partial fractions to evaluate ο„Έ 1 ( π‘₯ βˆ’ 1 ) π‘₯ 2 2 d .

  • A 1 4 | ( π‘₯ + 1 ) ( π‘₯ βˆ’ 1 ) | βˆ’ π‘₯ 2 ( π‘₯ βˆ’ 1 ) + 𝐾 l n 2
  • B 1 2 | | | π‘₯ + 1 π‘₯ βˆ’ 1 | | | βˆ’ π‘₯ 2 ( π‘₯ βˆ’ 1 ) + 𝐾 l n 2
  • C 1 4 | | | π‘₯ + 1 π‘₯ βˆ’ 1 | | | + π‘₯ 2 ( π‘₯ βˆ’ 1 ) + 𝐾 l n 2
  • D 1 4 | | | π‘₯ + 1 π‘₯ βˆ’ 1 | | | βˆ’ π‘₯ 2 ( π‘₯ βˆ’ 1 ) + 𝐾 l n 2
  • E 1 4 | | | π‘₯ + 1 π‘₯ βˆ’ 1 | | | βˆ’ 1 π‘₯ βˆ’ 1 + 𝐾 l n 2

Q6:

Use partial fractions to evaluate ο„Έ π‘₯ ( π‘₯ βˆ’ 1 ) ( π‘₯ + 2 π‘₯ + 1 ) π‘₯ 2 2 d .

  • A 1 2 | | ( π‘₯ βˆ’ 1 ) ( π‘₯ + 1 ) | | + 1 2 π‘₯ + 2 + 𝐾 l n 3
  • B 1 4 | | ( π‘₯ βˆ’ 1 ) ( π‘₯ + 1 ) | | βˆ’ 1 2 π‘₯ + 2 + 𝐾 l n 3
  • C 1 4 | | | ( π‘₯ βˆ’ 1 ) ( π‘₯ + 1 ) | | | + 1 2 π‘₯ + 2 + 𝐾 l n 3
  • D 1 4 | | ( π‘₯ βˆ’ 1 ) ( π‘₯ + 1 ) | | + 1 2 π‘₯ + 2 + 𝐾 l n 3
  • E 1 4 | | ( π‘₯ βˆ’ 1 ) ( π‘₯ + 1 ) | | + 1 π‘₯ + 1 + 𝐾 l n 3