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Page 1: Aim:  How do we integrate by partial fractions ? (I)

Aim: How do we integrate by partial fractions? (I)

Do Now: Write the partial fractions decomposition of

𝐴π‘₯βˆ’1

+𝐡π‘₯+2

=𝐴 (π‘₯+2 )+𝐡(π‘₯βˆ’ 1)

(π‘₯βˆ’1)(π‘₯+2)𝐴π‘₯+2𝐴+𝐡π‘₯βˆ’π΅=π‘₯+5

( 𝐴+𝐡 )π‘₯+(2 π΄βˆ’π΅ )=π‘₯+5

𝑨=𝟐 ,𝑩=βˆ’πŸ

Page 2: Aim:  How do we integrate by partial fractions ? (I)

∫ π‘₯+5

π‘₯2+π‘₯βˆ’ 2𝑑π‘₯

¿∫( 2π‘₯βˆ’1

βˆ’1

π‘₯+2 )𝑑π‘₯ΒΏ2 𝑙𝑛|π‘₯βˆ’ 1|βˆ’π‘™π‘›|π‘₯+2|+𝐢

𝑒=π‘₯βˆ’1 ,𝑑𝑒=𝑑π‘₯

𝑒=π‘₯+2,𝑑𝑒=𝑑π‘₯

Page 3: Aim:  How do we integrate by partial fractions ? (I)

A rational function whose numerator has higher degree than the denominator, divide and write the form of

∫ π‘₯3+π‘₯π‘₯βˆ’ 1

𝑑π‘₯¿∫(π‘₯2+π‘₯+2+2

π‘₯βˆ’1)𝑑π‘₯

ΒΏ π‘₯3

3+π‘₯

2

2+2π‘₯+2 𝑙𝑛|π‘₯βˆ’ 1|+𝐢

Page 4: Aim:  How do we integrate by partial fractions ? (I)

∫ π‘₯2+2 π‘₯βˆ’12 π‘₯3+3π‘₯2 βˆ’2π‘₯

𝑑π‘₯

¿∫ 12

1π‘₯

+15

12π‘₯βˆ’ 1

+110

1π‘₯+2

𝑑π‘₯

ΒΏ12𝑙𝑛|π‘₯|+ 1

10𝑙𝑛|2 π‘₯βˆ’1|βˆ’ 1

10𝑙𝑛|π‘₯+2|+𝐢

In integrating the middle term we have made the mental substitution u = 2x – 1, which gives du = 2dx and

Page 5: Aim:  How do we integrate by partial fractions ? (I)

∫ 𝑑π‘₯π‘₯2βˆ’π‘Ž2

ΒΏ 12π‘Žβˆ«( 1

π‘₯βˆ’π‘Žβˆ’

1π‘₯+π‘Ž )𝑑π‘₯

𝑨=πŸπŸπ’‚

,𝑩=βˆ’πŸπŸπ’‚

ΒΏ1

2π‘Ž(𝑙𝑛|π‘₯βˆ’π‘Ž|βˆ’π‘™π‘›|π‘₯+π‘Ž|)+𝐢

ΒΏ 12π‘Ž

𝑙𝑛|π‘₯βˆ’π‘Žπ‘₯+π‘Ž|+𝐢

Page 6: Aim:  How do we integrate by partial fractions ? (I)

∫ π‘₯4 βˆ’2π‘₯2+4 π‘₯+1π‘₯3βˆ’π‘₯2 βˆ’π‘₯+1

𝑑π‘₯ΒΏ 𝒙+𝟏+πŸ’ 𝒙

π’™πŸ‘βˆ’π’™πŸβˆ’ 𝒙+𝟏

π’™πŸ’βˆ’πŸπ’™πŸ+πŸ’ 𝒙+πŸπ’™πŸ‘βˆ’ π’™πŸβˆ’π’™+𝟏

¿∫ [π‘₯+1+ 1π‘₯βˆ’1

+ 2

(π‘₯βˆ’ 1)2βˆ’

1π‘₯+1 ]𝑑π‘₯  

ΒΏ π‘₯2

2+π‘₯+ 𝑙𝑛|π‘₯βˆ’1|βˆ’ 2

π‘₯βˆ’1βˆ’π‘™π‘›|π‘₯+1|+𝐢

ΒΏ π‘₯2

2+π‘₯βˆ’

2π‘₯βˆ’1

+𝑙𝑛|π‘₯βˆ’1π‘₯+1 |+𝐢


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