Worksheet: Integration by Partial Fractions with Linear Factors

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

Q1:

Use partial fractions to evaluate 𝑥+4𝑥+1(𝑥1)(𝑥+1)(𝑥+3)𝑥d.

  • A431|𝑥1|+121|𝑥+1|141|𝑥+3|+𝐾lnlnln
  • B34|𝑥1|+12|𝑥+1|14|𝑥+3|+𝐾lnlnln
  • C23|𝑥1|+12|𝑥+1|14|𝑥+3|+𝐾lnlnln
  • D43|𝑥1|+12|𝑥+1|+14|𝑥+3|+𝐾lnlnln
  • E43|𝑥1|+14|𝑥+1|14|𝑥+3|+𝐾lnlnln

Q2:

Use partial fractions to evaluate 6𝑥+7(𝑥+2)𝑥d.

  • A6|𝑥+2|+5|𝑥+2|+𝐾lnln
  • B6|𝑥+2|5(𝑥+2)+𝐾ln
  • C3|𝑥+2|4(𝑥+2)+𝐾ln
  • D6|𝑥+2|+5(𝑥+2)+𝐾ln
  • E6|𝑥+2|+5(𝑥+2)+𝐾ln

Q3:

Use partial fractions to evaluate 1(𝑥1)𝑥d.

  • A14|(𝑥+1)(𝑥1)|𝑥2(𝑥1)+𝐾ln
  • B14|||𝑥+1𝑥1|||𝑥2(𝑥1)+𝐾ln
  • C14|||𝑥+1𝑥1|||1𝑥1+𝐾ln
  • D12|||𝑥+1𝑥1|||𝑥2(𝑥1)+𝐾ln
  • E14|||𝑥+1𝑥1|||+𝑥2(𝑥1)+𝐾ln

Q4:

Use partial fractions to evaluate 𝑥(𝑥1)(𝑥+2𝑥+1)𝑥d.

  • A14||(𝑥1)(𝑥+1)||12𝑥+2+𝐾ln
  • B14||(𝑥1)(𝑥+1)||+12𝑥+2+𝐾ln
  • C12||(𝑥1)(𝑥+1)||+12𝑥+2+𝐾ln
  • D14|||(𝑥1)(𝑥+1)|||+12𝑥+2+𝐾ln
  • E14||(𝑥1)(𝑥+1)||+1𝑥+1+𝐾ln

Q5:

Use partial fractions to find an analytic expression for the integral 3𝑡9𝑡+8𝑡(𝑡2)𝑡.d

  • A2(𝑥)+(2𝑥)1𝑥2lnln
  • B2(𝑥)+(2𝑥)1𝑥21lnln
  • C2(𝑥)+(𝑥2)1𝑥21lnln
  • D2(𝑥)+(2𝑥)+1𝑥21lnln
  • E2(𝑥)+(𝑥2)+1𝑥21lnln

Q6:

Find 𝐹 so that 𝐹(𝑥)=𝑥(𝑥𝑒) and 𝐹(0)=2.

  • A𝐹(𝑥)=|(𝑒𝑥)|+𝑒𝑒𝑥ln
  • B𝐹(𝑥)=|(𝑥𝑒)|𝑥𝑒𝑥ln
  • C𝐹(𝑥)=|(𝑥𝑒)|+𝑒𝑒𝑥ln
  • D𝐹(𝑥)=|(𝑒𝑥)|𝑥𝑒𝑥ln
  • E𝐹(𝑥)=|(𝑒𝑥)|+𝑒𝑥𝑒ln

Q7:

Consider the function 𝑓(𝑥)=𝑥(𝑥𝑒) which is defined on {𝑒}.

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

  • A𝐹(𝑥)=(|𝑥𝑒|)+𝑒𝑒𝑥ln, 𝐹(2𝑒)=0.
  • B𝐹(𝑥)=(|𝑥𝑒|)𝑥𝑒𝑥ln, 𝐹(2𝑒)=1.
  • C𝐹(𝑥)=(|𝑥𝑒|)𝑒𝑒𝑥ln, 𝐹(2𝑒)=0.
  • D𝐹(𝑥)=(|𝑥𝑒|)𝑒𝑒𝑥ln, 𝐹(2𝑒)=2.
  • E𝐹(𝑥)=(|𝑥𝑒|)+𝑒𝑒𝑥ln, 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(𝑥)?

  • AYes, 𝐺(𝑥)=(𝑒𝑥)+𝑒𝑒𝑥𝑥<𝑒,(𝑥𝑒)+𝑒𝑒𝑥+1𝑥>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 𝐺𝐹 is not a constant function. Why is there no contradiction?

  • Abecause that result assumes that the domain is an interval
  • Bbecause 𝐺𝐹 is constant; it has the value 0 on (,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

Q8:

Use partial fractions to evaluate 𝑥+4(𝑥+6)(𝑥1)𝑥d.

  • A27|𝑥+6|57|𝑥1|+𝐾lnln
  • B271|𝑥6|+571|𝑥+1|+𝐾lnln
  • C27|𝑥+6|+57|𝑥1|+𝐾lnln
  • D27|𝑥1|+57|𝑥+6|+𝐾lnln
  • E27|𝑥+6|57|𝑥1|+𝐾lnln

Q9:

Use partial fractions to evaluate 1𝑥(𝑥+2)𝑥d.

  • A12|||𝑥𝑥+2|||+𝐶ln
  • Bln𝑥(𝑥+2)+𝐶
  • Cln|𝑥||𝑥+2|
  • Dln||𝑥𝑥+2||+𝐶
  • E12𝑥(𝑥+2)+𝐶ln

Q10:

Use partial fractions to evaluate 𝑥+4𝑥(𝑥+1)𝑥d.

  • A32+412+332lnlnln
  • B32+412+332lnlnln
  • C32412332lnlnln
  • D32412+332lnlnln
  • E32414+334lnlnln

Q11:

Use partial fractions to evaluate 1𝑡+𝑡2𝑡𝑡d.

  • A12|𝑡|16|𝑡+2|+13|𝑡1|+𝐾lnlnln
  • B12|𝑡|+16|𝑡+2|+13|𝑡1|+𝐾lnlnln
  • C12|𝑡|+16|𝑡+2|+13|𝑡1|+𝐾lnlnln
  • D12|𝑡|+16|𝑡2|+16|𝑡+1|+𝐾lnlnln
  • E12|𝑡|+16|𝑡+2|13|𝑡1|+𝐾lnlnln

Q12:

Use partial fractions to evaluate the integral 2𝑥(𝑥+7)𝑥d.

  • A2(𝑥7|𝑥+7|)+𝐾ln
  • B2(𝑥+7|𝑥+7|)+𝐾ln
  • C2|𝑥+7|+1𝑥+7+𝐾ln
  • D2𝑥+7+𝐾
  • E2|𝑥+7|+7𝑥+7+𝐾ln

Q13:

Use partial fractions to evaluate the integral 4𝑥5𝑥14𝑥d.

  • A49(|𝑥+2||𝑥7|)+𝐾lnln
  • B49(|𝑥7||𝑥+2|)+𝐾lnln
  • C49(|𝑥7|+|𝑥+2|)+𝐾lnln
  • D45(|𝑥7|+|𝑥+2|)+𝐾lnln
  • E94(|𝑥+7||𝑥2|)+𝐾lnln

Q14:

Use partial fractions to evaluate the integral 2𝑥+310𝑥+21𝑥+9𝑥𝑥d.

  • A13|||5+3𝑥|||+𝐾ln
  • B15|||5+3𝑥|||+𝐾ln
  • C13||𝑥5𝑥+3||+𝐾ln
  • D15||𝑥5𝑥+3||+𝐾ln
  • E13||𝑥2𝑥+3||+𝐾ln

Q15:

Use partial fractions to evaluate the integral 35𝑥20𝑥d.

  • A310(|𝑥2|+|𝑥+2|)+𝐾lnln
  • B310(|𝑥2||𝑥+2|)+𝐾lnln
  • C320(|𝑥2||𝑥+2|)+𝐾lnln
  • D310(|𝑥+2||𝑥2|)+𝐾lnln
  • E320(|𝑥+2||𝑥2|)+𝐾lnln

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