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Calculus and Integration by Parts Practice Questions

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Updated on Oct 1, 2011

To review these concepts, go to Calculus and Integration by Parts Study Guide.

Calculus and Integration by Parts Practice Questions

Evaluate the following integrals using integration by parts, substitution, or basic integration.

  1. x5ln(x)dx
  2. xsin(x)dx
  3. xsin(x2)dx
  4. (x + 3)cos(x)dx
  5. dx
  6.   x2sin(x)dx
  7. (x2 + sin(x))dx
  8. x2ex3 + 1 dx
  9. x2ex dx
  10. (x3 + 3x – 1)ln(x) dx
  11. (x + ln(x)) dx
  12. x – 1dx
  13. xx – 1dx
  14. xe–xdx
  15. xln(x) dx
  16. dx
  17. dx
  18. (x2 – 1)cos(x)dx
  19. dx
  20. sin(x)√cos(x)dx
  21. sin(x) · lncos(x)dx
  22. excos(x)dx

Solutions

  1. x6ln(x) – x6 + c, done by parts with u = ln(x)
  2. xcos(x) + sin(x) +c, by parts with u = x
  3. cos(x2) + c, by the substitution u = x2
  4. (x + 3)sin(x) + cos(x) + c, by parts with u = x + 3
  5. (ln(x))2 + c, by substituting u = ln(x)
  6. x2cos( x) + 2xsin(x) + 2cos(x) + c, using parts twice
  7. x3– cos(x) + c, by basic integration
  8. ex÷1 = c, by substituting u = x3 + 1
  9. x2ex – 2xex + 2ex + c, using parts twice
  10. (x4 + x2x)ln(x) – x4x2 + x + c, by parts with u = ln(x)
  11. x2 + xln(x) – x + c, evaluating, ∫ ln(x) dx by parts
  12. + c, substituting u = x – 1
  13. + c, by parts with u = x
  14. xexex + c, by parts with u = x
  15. xln(x) –x + c, by parts with u = x
  16. + c, by parts with u = ln(x)
  17. lnlxl + c, by basic integration
  18. (x2 – l)sin(x) + 2xcos(x) – 2sin(x) + c, using parts twice
  19. + c, by substituting u =
  20. , by substituting u = cos(x)
  21. –cos(x) · ln(cos(x)) + cos(x) + c, by parts with u = ln( cos(x)) . This could also be solved by substitution with u = cos(x) , though it would require knowing ln(u) du.
  22. (exsin(x) + excos(x)) + c, by parts twice, plus the trick from the previous example
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