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Lesson Worksheet: Integration by Partial Fractions with Linear Factors Mathematics • Higher Education

In this worksheet, we will practice using partial fractions to evaluate integrals of rational functions with linear factors.


Use partial fractions to evaluate ο„Έπ‘₯+4π‘₯+1(π‘₯βˆ’1)(π‘₯+1)(π‘₯+3)π‘₯d.

  • A431|π‘₯βˆ’1|+121|π‘₯+1|βˆ’141|π‘₯+3|+𝐾lnlnln
  • B34|π‘₯βˆ’1|+12|π‘₯+1|βˆ’14|π‘₯+3|+𝐾lnlnln
  • C23|π‘₯βˆ’1|+12|π‘₯+1|βˆ’14|π‘₯+3|+𝐾lnlnln
  • D43|π‘₯βˆ’1|+12|π‘₯+1|+14|π‘₯+3|+𝐾lnlnln
  • E43|π‘₯βˆ’1|+14|π‘₯+1|βˆ’14|π‘₯+3|+𝐾lnlnln


Use partial fractions to evaluate ο„Έ6π‘₯+7(π‘₯+2)π‘₯d.

  • A6|π‘₯+2|+5|π‘₯+2|+𝐾lnln
  • B6|π‘₯+2|βˆ’5(π‘₯+2)+𝐾ln
  • C3|π‘₯+2|βˆ’4(π‘₯+2)+𝐾ln
  • D6|π‘₯+2|+5(π‘₯+2)+𝐾ln
  • E6|π‘₯+2|+5(π‘₯+2)+𝐾ln


Use partial fractions to evaluate ο„Έ1(π‘₯βˆ’1)π‘₯d.

  • A14|(π‘₯+1)(π‘₯βˆ’1)|βˆ’π‘₯2(π‘₯βˆ’1)+𝐾ln
  • B14|||π‘₯+1π‘₯βˆ’1|||βˆ’π‘₯2(π‘₯βˆ’1)+𝐾ln
  • C14|||π‘₯+1π‘₯βˆ’1|||βˆ’1π‘₯βˆ’1+𝐾ln
  • D12|||π‘₯+1π‘₯βˆ’1|||βˆ’π‘₯2(π‘₯βˆ’1)+𝐾ln
  • E14|||π‘₯+1π‘₯βˆ’1|||+π‘₯2(π‘₯βˆ’1)+𝐾ln


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

  • A14||(π‘₯βˆ’1)(π‘₯+1)||βˆ’12π‘₯+2+𝐾ln
  • B14||(π‘₯βˆ’1)(π‘₯+1)||+12π‘₯+2+𝐾ln
  • C12||(π‘₯βˆ’1)(π‘₯+1)||+12π‘₯+2+𝐾ln
  • D14|||(π‘₯βˆ’1)(π‘₯+1)|||+12π‘₯+2+𝐾ln
  • E14||(π‘₯βˆ’1)(π‘₯+1)||+1π‘₯+1+𝐾ln


Use partial fractions to find an analytic expression for the integral ο„Έ3π‘‘βˆ’9𝑑+8𝑑(π‘‘βˆ’2)𝑑,π‘₯β‰₯1.ο—οŠ§οŠ¨οŠ¨dwhere

  • A2(π‘₯)+(2βˆ’π‘₯)βˆ’1π‘₯βˆ’2lnln
  • B2(π‘₯)+(2βˆ’π‘₯)βˆ’1π‘₯βˆ’2βˆ’1lnln
  • C2(π‘₯)+(π‘₯βˆ’2)βˆ’1π‘₯βˆ’2βˆ’1lnln
  • D2(π‘₯)+(2βˆ’π‘₯)+1π‘₯βˆ’2βˆ’1lnln
  • E2(π‘₯)+(π‘₯βˆ’2)+1π‘₯βˆ’2βˆ’1lnln


Find 𝐹 so that 𝐹′(π‘₯)=π‘₯(π‘₯βˆ’π‘’) and 𝐹(0)=2.

  • A𝐹(π‘₯)=|(π‘’βˆ’π‘₯)|+π‘’π‘’βˆ’π‘₯ln
  • B𝐹(π‘₯)=|(π‘₯βˆ’π‘’)|βˆ’π‘₯π‘’βˆ’π‘₯ln
  • C𝐹(π‘₯)=|(π‘₯βˆ’π‘’)|+π‘’π‘’βˆ’π‘₯ln
  • D𝐹(π‘₯)=|(π‘’βˆ’π‘₯)|βˆ’π‘₯π‘’βˆ’π‘₯ln
  • E𝐹(π‘₯)=|(π‘’βˆ’π‘₯)|+𝑒π‘₯βˆ’π‘’ln


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

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

  • A𝐹(π‘₯)=(|π‘₯βˆ’π‘’|)+π‘’π‘’βˆ’π‘₯ln, 𝐹(2𝑒)=0.
  • B𝐹(π‘₯)=(|π‘₯βˆ’π‘’|)βˆ’π‘₯π‘’βˆ’π‘₯ln, 𝐹(2𝑒)=βˆ’1.
  • C𝐹(π‘₯)=(|π‘₯βˆ’π‘’|)βˆ’π‘’π‘’βˆ’π‘₯ln, 𝐹(2𝑒)=0.
  • D𝐹(π‘₯)=(|π‘₯βˆ’π‘’|)βˆ’π‘’π‘’βˆ’π‘₯ln, 𝐹(2𝑒)=2.
  • E𝐹(π‘₯)=(|π‘₯βˆ’π‘’|)+π‘’π‘’βˆ’π‘₯ln, 𝐹(2𝑒)=2.

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

  • AYes, 𝐺(π‘₯)=(π‘’βˆ’π‘₯)+π‘’π‘’βˆ’π‘₯π‘₯<𝑒,(π‘₯βˆ’π‘’)+π‘’π‘’βˆ’π‘₯+1π‘₯>𝑒.lniflnif
  • BNo

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

  • ABecause that result assumes that the domain is an interval
  • BBecause πΊβˆ’πΉ is constant; it has the value 0 on (βˆ’βˆž,𝑒) and the value 1 on (𝑒,∞)
  • CBecause that result only holds sometimes; sometimes it fails
  • DBecause neither 𝐹 nor 𝐺 is differentiable; that result only applies to differentiable functions
  • EBecause that result requires additional conditions to the antiderivatives


Use partial fractions to evaluate ο„Έπ‘₯+4(π‘₯+6)(π‘₯βˆ’1)π‘₯d.

  • A27|π‘₯+6|βˆ’57|π‘₯βˆ’1|+𝐾lnln
  • B271|π‘₯βˆ’6|+571|π‘₯+1|+𝐾lnln
  • C27|π‘₯+6|+57|π‘₯βˆ’1|+𝐾lnln
  • D27|π‘₯βˆ’1|+57|π‘₯+6|+𝐾lnln
  • Eβˆ’27|π‘₯+6|βˆ’57|π‘₯βˆ’1|+𝐾lnln


Use partial fractions to evaluate ο„Έ1π‘₯(π‘₯+2)π‘₯d.

  • A12|||ο„žπ‘₯π‘₯+2|||+𝐢ln
  • Bln√π‘₯(π‘₯+2)+𝐢
  • Clnο„Ÿ|π‘₯||π‘₯+2|
  • Dln||π‘₯π‘₯+2||+𝐢
  • E12√π‘₯(π‘₯+2)+𝐢ln


Use partial fractions to evaluate ο„Έπ‘₯+4π‘₯(π‘₯+1)π‘₯d.

  • A32+412+332lnlnln
  • Bβˆ’32+412+332lnlnln
  • Cβˆ’32βˆ’412βˆ’332lnlnln
  • Dβˆ’32βˆ’412+332lnlnln
  • Eβˆ’32βˆ’414+334lnlnln

This lesson includes 8 additional questions and 63 additional question variations for subscribers.

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