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.

  • A 4 3 1 | 𝑥 1 | + 1 2 1 | 𝑥 + 1 | 1 4 1 | 𝑥 + 3 | + 𝐾 l n l n l n
  • B 3 4 | 𝑥 1 | + 1 2 | 𝑥 + 1 | 1 4 | 𝑥 + 3 | + 𝐾 l n l n l n
  • C 2 3 | 𝑥 1 | + 1 2 | 𝑥 + 1 | 1 4 | 𝑥 + 3 | + 𝐾 l n l n l n
  • D 4 3 | 𝑥 1 | + 1 2 | 𝑥 + 1 | + 1 4 | 𝑥 + 3 | + 𝐾 l n l n l n
  • E 4 3 | 𝑥 1 | + 1 4 | 𝑥 + 1 | 1 4 | 𝑥 + 3 | + 𝐾 l n l n l n

Q2:

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

  • A 6 | 𝑥 + 2 | + 5 | 𝑥 + 2 | + 𝐾 l n l n
  • B 6 | 𝑥 + 2 | 5 ( 𝑥 + 2 ) + 𝐾 l n
  • C 3 | 𝑥 + 2 | 4 ( 𝑥 + 2 ) + 𝐾 l n
  • D 6 | 𝑥 + 2 | + 5 ( 𝑥 + 2 ) + 𝐾 l n
  • E 6 | 𝑥 + 2 | + 5 ( 𝑥 + 2 ) + 𝐾 l n

Q3:

Use partial fractions to evaluate 1(𝑥1)𝑥d.

  • A 1 4 | ( 𝑥 + 1 ) ( 𝑥 1 ) | 𝑥 2 ( 𝑥 1 ) + 𝐾 l n
  • B 1 4 | | | 𝑥 + 1 𝑥 1 | | | 𝑥 2 ( 𝑥 1 ) + 𝐾 l n
  • C 1 4 | | | 𝑥 + 1 𝑥 1 | | | 1 𝑥 1 + 𝐾 l n
  • D 1 2 | | | 𝑥 + 1 𝑥 1 | | | 𝑥 2 ( 𝑥 1 ) + 𝐾 l n
  • E 1 4 | | | 𝑥 + 1 𝑥 1 | | | + 𝑥 2 ( 𝑥 1 ) + 𝐾 l n

Q4:

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

  • A 1 4 | | ( 𝑥 1 ) ( 𝑥 + 1 ) | | 1 2 𝑥 + 2 + 𝐾 l n
  • B 1 4 | | ( 𝑥 1 ) ( 𝑥 + 1 ) | | + 1 2 𝑥 + 2 + 𝐾 l n
  • C 1 2 | | ( 𝑥 1 ) ( 𝑥 + 1 ) | | + 1 2 𝑥 + 2 + 𝐾 l n
  • D 1 4 | | | ( 𝑥 1 ) ( 𝑥 + 1 ) | | | + 1 2 𝑥 + 2 + 𝐾 l n
  • E 1 4 | | ( 𝑥 1 ) ( 𝑥 + 1 ) | | + 1 𝑥 + 1 + 𝐾 l n

Q5:

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

  • A 2 ( 𝑥 ) + ( 2 𝑥 ) 1 𝑥 2 l n l n
  • B 2 ( 𝑥 ) + ( 2 𝑥 ) 1 𝑥 2 1 l n l n
  • C 2 ( 𝑥 ) + ( 𝑥 2 ) 1 𝑥 2 1 l n l n
  • D 2 ( 𝑥 ) + ( 2 𝑥 ) + 1 𝑥 2 1 l n l n
  • E 2 ( 𝑥 ) + ( 𝑥 2 ) + 1 𝑥 2 1 l n l n

Q6:

Find 𝐹 so that 𝐹(𝑥)=𝑥(𝑥𝑒) and 𝐹(0)=2.

  • A 𝐹 ( 𝑥 ) = | ( 𝑒 𝑥 ) | + 𝑒 𝑒 𝑥 l n
  • B 𝐹 ( 𝑥 ) = | ( 𝑥 𝑒 ) | 𝑥 𝑒 𝑥 l n
  • C 𝐹 ( 𝑥 ) = | ( 𝑥 𝑒 ) | + 𝑒 𝑒 𝑥 l n
  • D 𝐹 ( 𝑥 ) = | ( 𝑒 𝑥 ) | 𝑥 𝑒 𝑥 l n
  • E 𝐹 ( 𝑥 ) = | ( 𝑒 𝑥 ) | + 𝑒 𝑥 𝑒 l n

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)?

  • A 𝐹 ( 𝑥 ) = ( | 𝑥 𝑒 | ) + 𝑒 𝑒 𝑥 l n , 𝐹 ( 2 𝑒 ) = 0 .
  • B 𝐹 ( 𝑥 ) = ( | 𝑥 𝑒 | ) 𝑥 𝑒 𝑥 l n , 𝐹 ( 2 𝑒 ) = 1 .
  • C 𝐹 ( 𝑥 ) = ( | 𝑥 𝑒 | ) 𝑒 𝑒 𝑥 l n , 𝐹 ( 2 𝑒 ) = 0 .
  • D 𝐹 ( 𝑥 ) = ( | 𝑥 𝑒 | ) 𝑒 𝑒 𝑥 l n , 𝐹 ( 2 𝑒 ) = 2 .
  • E 𝐹 ( 𝑥 ) = ( | 𝑥 𝑒 | ) + 𝑒 𝑒 𝑥 l n , 𝐹 ( 2 𝑒 ) = 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.

  • A 2 7 | 𝑥 + 6 | 5 7 | 𝑥 1 | + 𝐾 l n l n
  • B 2 7 1 | 𝑥 6 | + 5 7 1 | 𝑥 + 1 | + 𝐾 l n l n
  • C 2 7 | 𝑥 + 6 | + 5 7 | 𝑥 1 | + 𝐾 l n l n
  • D 2 7 | 𝑥 1 | + 5 7 | 𝑥 + 6 | + 𝐾 l n l n
  • E 2 7 | 𝑥 + 6 | 5 7 | 𝑥 1 | + 𝐾 l n l n

Q9:

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

  • A 1 2 | | | 𝑥 𝑥 + 2 | | | + 𝐶 l n
  • B l n 𝑥 ( 𝑥 + 2 ) + 𝐶
  • C l n | 𝑥 | | 𝑥 + 2 |
  • D l n | | 𝑥 𝑥 + 2 | | + 𝐶
  • E 1 2 𝑥 ( 𝑥 + 2 ) + 𝐶 l n

Q10:

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

  • A 3 2 + 4 1 2 + 3 3 2 l n l n l n
  • B 3 2 + 4 1 2 + 3 3 2 l n l n l n
  • C 3 2 4 1 2 3 3 2 l n l n l n
  • D 3 2 4 1 2 + 3 3 2 l n l n l n
  • E 3 2 4 1 4 + 3 3 4 l n l n l n

Q11:

Use partial fractions to evaluate 1𝑡+𝑡2𝑡𝑡d.

  • A 1 2 | 𝑡 | 1 6 | 𝑡 + 2 | + 1 3 | 𝑡 1 | + 𝐾 l n l n l n
  • B 1 2 | 𝑡 | + 1 6 | 𝑡 + 2 | + 1 3 | 𝑡 1 | + 𝐾 l n l n l n
  • C 1 2 | 𝑡 | + 1 6 | 𝑡 + 2 | + 1 3 | 𝑡 1 | + 𝐾 l n l n l n
  • D 1 2 | 𝑡 | + 1 6 | 𝑡 2 | + 1 6 | 𝑡 + 1 | + 𝐾 l n l n l n
  • E 1 2 | 𝑡 | + 1 6 | 𝑡 + 2 | 1 3 | 𝑡 1 | + 𝐾 l n l n l n

Q12:

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

  • A 2 ( 𝑥 7 | 𝑥 + 7 | ) + 𝐾 l n
  • B 2 ( 𝑥 + 7 | 𝑥 + 7 | ) + 𝐾 l n
  • C 2 | 𝑥 + 7 | + 1 𝑥 + 7 + 𝐾 l n
  • D 2 𝑥 + 7 + 𝐾
  • E 2 | 𝑥 + 7 | + 7 𝑥 + 7 + 𝐾 l n

Q13:

Use partial fractions to evaluate the integral 4𝑥5𝑥14𝑥d.

  • A 4 9 ( | 𝑥 + 2 | | 𝑥 7 | ) + 𝐾 l n l n
  • B 4 9 ( | 𝑥 7 | | 𝑥 + 2 | ) + 𝐾 l n l n
  • C 4 9 ( | 𝑥 7 | + | 𝑥 + 2 | ) + 𝐾 l n l n
  • D 4 5 ( | 𝑥 7 | + | 𝑥 + 2 | ) + 𝐾 l n l n
  • E 9 4 ( | 𝑥 + 7 | | 𝑥 2 | ) + 𝐾 l n l n

Q14:

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

  • A 1 3 | | | 5 + 3 𝑥 | | | + 𝐾 l n
  • B 1 5 | | | 5 + 3 𝑥 | | | + 𝐾 l n
  • C 1 3 | | 𝑥 5 𝑥 + 3 | | + 𝐾 l n
  • D 1 5 | | 𝑥 5 𝑥 + 3 | | + 𝐾 l n
  • E 1 3 | | 𝑥 2 𝑥 + 3 | | + 𝐾 l n

Q15:

Use partial fractions to evaluate the integral 35𝑥20𝑥d.

  • A 3 1 0 ( | 𝑥 2 | + | 𝑥 + 2 | ) + 𝐾 l n l n
  • B 3 1 0 ( | 𝑥 2 | | 𝑥 + 2 | ) + 𝐾 l n l n
  • C 3 2 0 ( | 𝑥 2 | | 𝑥 + 2 | ) + 𝐾 l n l n
  • D 3 1 0 ( | 𝑥 + 2 | | 𝑥 2 | ) + 𝐾 l n l n
  • E 3 2 0 ( | 𝑥 + 2 | | 𝑥 2 | ) + 𝐾 l n l n

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