Lesson Worksheet: Integration by Parts Mathematics • Higher Education

In this worksheet, we will practice using integration by parts to find the integral of a product of functions.

Q1:

Use integration by parts to evaluate ο„Έπ‘₯π‘₯π‘₯sind.

  • Aπ‘₯π‘₯βˆ’π‘₯+cossinC
  • Bπ‘₯(π‘₯βˆ’π‘₯)+sincosC
  • CsincosCπ‘₯+π‘₯π‘₯+
  • Dβˆ’π‘₯βˆ’π‘₯π‘₯+sincosC
  • EsincosCπ‘₯βˆ’π‘₯π‘₯+

Q2:

Determine ο„Έ9π‘₯+7𝑒π‘₯οŠ«ο—d.

  • Aβˆ’95ο€Όπ‘₯+4445οˆπ‘’+οŠ±οŠ«ο—C
  • B95ο€Όπ‘₯+4445οˆπ‘’+οŠ±οŠ«ο—C
  • Cβˆ’15(9π‘₯+7)𝑒+οŠ±οŠ«ο—C
  • Dβˆ’95ο€Όπ‘₯+2645οˆπ‘’+οŠ±οŠ«ο—C
  • Eβˆ’9ο€Όπ‘₯+4445οˆπ‘’+οŠ±οŠ«ο—C

Q3:

Determine ο„Έ(3π‘₯+4)𝑒π‘₯οŠ¨ο—d.

  • A𝑒92π‘₯+6π‘₯+10+ο—οŠ¨C
  • B𝑒92π‘₯+3π‘₯+1+ο—οŠ¨C
  • C𝑒9π‘₯+3π‘₯+10+ο—οŠ¨C
  • D𝑒9π‘₯+6π‘₯+10+ο—οŠ¨C

Q4:

Evaluate ο„Έπ‘₯𝑒π‘₯οŠ§οŠ¦οŠ¨ο—d.

  • Aπ‘’βˆ’2
  • B1βˆ’π‘’
  • C𝑒+2
  • D2βˆ’3𝑒
  • Eπ‘’βˆ’1

Q5:

Integrate ο„Έπ‘₯π‘₯lnd by parts using 𝑒=π‘₯ln and dd𝑣=π‘₯.

  • AlnCπ‘₯βˆ’π‘₯+
  • Bπ‘₯π‘₯+1+lnC
  • Cπ‘₯(π‘₯+1)+lnC
  • Dπ‘₯(π‘₯βˆ’1)+lnC
  • Eπ‘₯π‘₯βˆ’1+lnC

Q6:

Determine ο„Έπ‘₯π‘₯π‘₯lnd.

  • A19π‘₯(3π‘₯βˆ’1)+lnC
  • B13π‘₯(3π‘₯βˆ’1)+lnC
  • C19π‘₯(3π‘₯+1)+lnC
  • D13π‘₯(3π‘₯+1)+lnC
  • E19π‘₯(π‘₯βˆ’1)+lnC

Q7:

Calculate ο„Έ(π‘₯)π‘₯tand.

  • Aπœ‹2βˆ’(2)ln
  • Bπœ‹4+(2)2ln
  • Cπœ‹4βˆ’(2)2ln
  • Dπœ‹2
  • Eπœ‹4βˆ’(2)ln

Q8:

By setting 𝑒=𝑒 and dcosd𝑣=π‘₯π‘₯, evaluate 𝑒π‘₯π‘₯cosd by integrating by parts.

  • A2𝑒(π‘₯+π‘₯)+sincosC
  • B12𝑒(π‘₯βˆ’π‘₯)+sincosC
  • C𝑒(π‘₯+π‘₯)+sincosC
  • D2𝑒(π‘₯βˆ’π‘₯)+sincosC
  • E12𝑒(π‘₯+π‘₯)+sincosC

Q9:

A curve passes through ο€Ό0,715 and the tangent at its point (π‘₯,𝑦) has slope 8π‘₯√2π‘₯+1. What is the equation of the curve?

  • A𝑦=815(2π‘₯+1)(3π‘₯βˆ’1)+1
  • B𝑦=815(2π‘₯+1)(3π‘₯βˆ’1)βˆ’115
  • C𝑦=415(2π‘₯+1)(8π‘₯βˆ’1)+1115
  • D𝑦=815(2π‘₯+1)(3π‘₯βˆ’1)βˆ’1615

Q10:

Determine ο„Έ2𝑒π‘₯3(π‘₯+1)π‘₯ο—οŠ¨d.

  • A2𝑒3(π‘₯+1)+C
  • B2𝑒(2π‘₯+1)3(π‘₯+1)+C
  • Cβˆ’2𝑒(2π‘₯+1)3(π‘₯+1)+C
  • Dβˆ’2𝑒3(π‘₯+1)+C

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