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Integration By Partial Fractions

  • Integrating by partial fractions:
    • Modify denominator into factored and multiple terms
    • Multiply constants A, B, C… into corresponding denominator terms
    • Rearrange terms to match original numerator
    • Set-up system of equations
    • Solve for all constants
    • Integrate using fundamental integral rules
  • Know the process very well, as this rarely changes for the AP exam

Integration By Partial Fractions

Expand [4/((x + 1)(x + 2))] using Partial Fractions
  • Setup form
  • [4/((x + 1)(x + 2))] = [A/(x + 1)] + [B/(x + 2)]
  • Multiply each side by the common demoninator
  • (x + 1)(x + 2)[4/((x + 1)(x + 2))] = (x + 1)(x + 2)( [A/(x + 1)] + [B/(x + 2)] )
  • 4 = A(x + 2) + B(x + 1)
  • Set x = − 1 to solve forA
  • 4 = A( − 1 + 2) + B( − 1 + 1)
  • 4 = A
  • Set x = − 2 to solve for B
  • 4 = A( − 2 + 2) + B( − 2 + 1)
  • 4 = − B
  • − 4 = B
Solve partial fractions
[4/((x + 1)(x + 2))] = [A/(x + 1)] + [B/(x + 2)] = [4/(x + 1)] − [4/(x + 2)]
Solve ∫ [4/((x + 1)(x + 2))] dx
  • Use partial fractions to integrate
  • [4/((x + 1)(x + 2))] dx = ∫ [4/(x + 1)] − [4/(x + 2)] dx
  • = ∫ [4/(x + 1)] dx − ∫ [4/(x + 2)] dx
  • Use Chain Rule to integrate
  • [4/(x + 1)] dx = 4ln|x + 1| + C
[4/(x + 2)] dx = 4ln|x + 2| + C∫ [4/(x + 1)] dx − ∫ [4/(x + 2)] dx = 4ln|x + 1| − 4ln|x + 2| + C
0e [32/((x + 1)(x + 2))] dx =
  • Note the factors of 32
  • 0e [32/((x + 1)(x + 2))] dx = 8∫0e [4/((x + 1)(x + 2))] dx
  • = 8[ 4ln|x + 1| − 4ln|x + 2| ]0e
  • = 8(4ln1 − 4ln2 − 4ln(1 + e) + 4ln(2 + e))
= 32( − ln2 − 2ln( (1 + e)(2 + e) )
Expand [(x − 9)/(x2 − 10x + 21)] using Partial Fractions
  • Simplify the denominator
  • [(x − 9)/(x2 − 10x + 21)] = [(x − 9)/((x − 7)(x − 3))]
  • Setup form
  • [(x − 9)/((x − 7)(x − 3))] = [A/(x − 7)] + [B/(x − 3)]
  • Multiply each side by the common demoninator
  • (x − 7)(x − 3)[(x − 9)/((x − 7)(x − 3))] = (x − 7)(x − 3)( [A/(x − 7)] + [B/(x − 3)] )
  • x − 9 = A(x − 3) + B(x − 7)
  • Set x = 7 to solve for A
  • 7 − 9 = A(7 − 3) + B(7 − 7)
  • − 2 = 4A
  • − [1/2] = A
  • Set x = 3 to solve for B
  • 3 − 9 = A(3 − 3) + B(3 − 7)
  • − 6 = − 4B
  • [2/3] = B
Solve partion fractions
[(x − 9)/((x − 7)(x − 3))] = [A/(x − 7)] + [B/(x − 3)] = [( − [1/2])/(x − 7)] + [([2/3])/(x − 3)] = [2/3]( [1/(x − 3)] ) − [1/2]( [1/(x − 7)] )
Solve ∫ [(x − 9)/(x2 − 10x + 21)] dx
  • Use partial fractions to integrate
  • [(x − 9)/(x2 − 10x + 21)] dx = ∫ [2/3]( [1/(x − 3)] ) − [1/2]( [1/(x − 7)] ) dx
  • = ∫ [2/3]( [1/(x − 3)] ) dx − ∫ [1/2]( [1/(x − 7)] ) dx
  • = [2/3]∫ ( [1/(x − 3)] ) dx − [1/2]∫ ( [1/(x − 7)] ) dx
  • Use Chain Rule to integrate
  • [2/3]∫ ( [1/(x − 3)] ) dx = [2/3]ln|x − 3| + C
  • [1/2]∫ ( [1/(x − 7)] ) dx = [1/2]ln|x − 7| + C
[2/3]∫ ( [1/(x − 3)] ) dx − [1/2]∫ ( [1/(x − 7)] ) dx = [2/3]ln|x − 3| + C − [1/2]ln|x − 7| + C
Expand [2x/(x2 + 2x − 8)] using Partial Fractions
  • Simplify the denominator
  • [2x/(x2 + 2x − 8)] = [2x/((x + 4)(x − 2))]
  • Setup form
  • [2x/((x + 4)(x − 2))] = [A/(x + 4)] + [B/(x − 2)]
  • Multiply each side by the common demoninator
  • (x + 4)(x − 2)[2x/((x + 4)(x − 2))] = (x + 4)(x − 2)( [A/(x + 4)] + [B/(x − 2)] )
  • 2x = A(x − 2) + B(x + 4)
  • Set x = − 4 to solve for A
  • 2( − 4) = A( − 4 − 2) + B( − 4 + 4)
  • − 8 = − 6A
  • [4/3] = A
  • Set x = 2 to solve for B
  • 2(2) = A(2 − 2) + B(2 + 4)
  • 4 = 6B
  • [2/3] = B
  • Solve partial fractions
  • [2x/((x + 4)(x − 2))] = [A/(x + 4)] + [B/(x − 2)]
  • = [([4/3])/(x + 4)] + [([2/3])/(x − 2)]
= [4/3]( [1/(x + 4)] ) + [2/3]( [1/(x − 2)] )
Solve ∫ [2x/(x2 + 2x − 8)] dx
  • Use partial fractions to integrate
  • [2x/(x2 + 2x − 8)] dx = ∫ [4/3]( [1/(x + 4)] ) + [2/3]( [1/(x − 2)] ) dx
  • = ∫ [4/3]( [1/(x + 4)] ) dx + ∫ [2/3]( [1/(x − 2)] ) dx
  • = [4/3]∫ ( [1/(x + 4)] ) dx + [2/3]∫ ( [1/(x − 2)] ) dx
  • Use Chain Rule to integrate
  • [4/3]∫ ( [1/(x + 4)] ) dx = [4/3]ln|x + 3| + C
  • [2/3]∫ ( [1/(x − 2)] ) dx = [1/2]ln|x − 2| + C
[4/3]∫ ( [1/(x + 4)] ) dx + [2/3]∫ ( [1/(x − 2)] ) dx = [4/3]ln|x + 4| + [2/3]ln|x − 2| + C
Expand [1/(x2 − 3x)] using Partial Fractions
  • Simplify the denominator
  • [1/(x2 − 3x)] = [1/((x)(x − 3))]
  • Setup form
  • [1/((x)(x − 3))] = [A/x] + [B/(x − 3)]
  • Multiply each side by the common demoninator
  • (x)(x − 3)[1/((x)(x − 3))] = (x)(x − 3)( [A/x] + [B/(x − 3)] )
  • 1 = A(x) + B(x − 3)
  • Set x = 3 to solve for A
  • 1 = A(3) + B(3 − 3)
  • 1 = 3A
  • [1/3] = A
  • Set x = 0 to solve for B
  • 1 = A(0) + B(0 − 3)
  • 1 = − 3B
  • − [1/3] = B
[1/((x)(x − 3))] = [A/x] + [B/(x − 3)] = [([1/3])/x] + [( − [1/3])/(x − 3)] = [1/3]( [1/x] ) − [1/3]( [1/(x − 3)] )
Setup ∫ [lnx/((x − 3)2)]dx into Integration by Parts
  • Let u = lnx, dv = [1/((x − 3)2)]
  • du = [dx/x]
  • v = [( − 1)/(x − 3)] + C
Setup Integration by Parts
[lnx/((x − 3)2)]dx = lnx( [( − 1)/(x − 3)] ) − ∫ [( − 1)/(x − 3)]( [dx/x] ) + C
Solve ∫ [lnx/((x − 3)2)] dx
  • Apply Integration by Parts
  • [lnx/((x − 3)2)] = lnx( [( − 1)/(x − 3)] ) − ∫ [( − 1)/(x − 3)]( [dx/x] ) + C
  • Apply Partial Fractions
  • lnx( [( − 1)/(x − 3)] ) − ∫ [( − 1)/(x − 3)]( [dx/x] ) + C
  • = lnx( [( − 1)/(x − 3)] ) + ∫ [dx/((x)(x − 3))] + C
  • = − [lnx/(x − 3)] + ∫ [1/3]( [1/x] ) − [1/3]( [1/(x − 3)] ) + C
  • = − [lnx/(x − 3)] + [1/3]∫ [1/x]dx − [1/3]∫ [1/(x − 3)]dx + C
= − [lnx/(x − 3)] + [1/3]ln|x| − [1/3]ln|x − 3| + C

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Integration By Partial Fractions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Integration by Partial Fractions: Goal 0:08
  • Example 1 2:10
  • Example 2 7:10
  • Example 3 10:06
  • Example 4 14:23
  • Example 5 16:26