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Aim: How do we integrate by partial fractions? (II) Do Now: Write the partial fraction decomposition of 2 2 βˆ’ + 4 ( 2 + 4 ) ΒΏ + + 2 + 4 2 2 βˆ’ + 4 = ( 2 + 4 ) + ( + ) + =2 , =βˆ’ 1 , 4 =4 =1 , = 1 , =βˆ’ 1 2 2 βˆ’ + 4 3 + 4 = 1 + βˆ’ 1 2 + 4

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

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Aim: How do we integrate by partial fractions? (II). Do Now: Write the partial fraction decomposition of. =. =. ,. , d u = 2 x d x. ,. ,. ,. - PowerPoint PPT Presentation

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

Aim: How do we integrate by partial fractions? (II) Do Now: Write the partial fraction decomposition of

2π‘₯2βˆ’π‘₯+4π‘₯ (π‘₯2+4)

¿𝐴π‘₯

+𝐡π‘₯+𝐢π‘₯2+4

2 π‘₯2βˆ’π‘₯+4=𝐴 (π‘₯2+4 )+ (𝐡π‘₯+𝐢 ) π‘₯

𝐴+𝐡=2 ,𝐢=βˆ’1 ,4 𝐴=4 𝐴=1 ,𝐡=1,𝐢=βˆ’1

2π‘₯2βˆ’π‘₯+4π‘₯3+4 π‘₯

=1π‘₯

+ π‘₯βˆ’1π‘₯2+4

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

∫ 2 π‘₯2βˆ’π‘₯+4π‘₯3+4 π‘₯

𝑑π‘₯¿∫(1π‘₯

+π‘₯βˆ’1

π‘₯2+4)𝑑π‘₯

¿∫ 1π‘₯𝑑π‘₯+∫ π‘₯βˆ’1

π‘₯2+4𝑑π‘₯

¿∫ 1π‘₯𝑑π‘₯+∫ π‘₯

π‘₯2+4𝑑π‘₯βˆ’βˆ« 1

π‘₯2+4𝑑π‘₯

𝒖=π’™πŸ+πŸ’

𝒅𝒖=𝟐 𝒙 𝒅𝒙¿ 𝑙𝑛|π‘₯|+ 12𝑙𝑛|π‘₯2+4|βˆ’ 1

2π‘‘π‘Žπ‘›βˆ’ 1( π‘₯2 )+𝐢

∫ π’…π’™π’™πŸ+π’‚πŸ

=πŸπ’‚π’•π’‚π’βˆ’πŸ( 𝒙𝒂 )+π‘ͺ

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

∫ 4 π‘₯2βˆ’3 π‘₯+24 π‘₯2βˆ’4 π‘₯+3

𝑑π‘₯¿∫1+π‘₯βˆ’1

4 π‘₯2βˆ’4 π‘₯+3𝑑π‘₯

¿∫1𝑑π‘₯+∫ π‘₯βˆ’1

(2 π‘₯βˆ’1 )2+2  π‘‘π‘₯

πŸ’π’™πŸβˆ’πŸ’ 𝒙+πŸ‘=(𝟐 π’™βˆ’πŸ )𝟐+𝟐 𝒖=𝟐 π’™βˆ’πŸ ,𝒅𝒖=πŸπ’…π’™π’™=

𝟏𝟐

(𝒖+𝟏)

¿∫1𝑑π‘₯+12∫

12

(𝑒+1 )βˆ’1

𝑒2+2𝑑𝑒¿ π‘₯+

14∫

π‘’βˆ’1

𝑒2+2𝑑𝑒

=

ΒΏ π‘₯+18𝑙𝑛|𝑒2+2|βˆ’ 1

4βˆ™

1

√2π‘‘π‘Žπ‘›βˆ’1( π‘’βˆšπ‘Ž )

ΒΏ π‘₯+18𝑙𝑛|4 π‘₯2βˆ’4 π‘₯+3|βˆ’ 1

4 √2π‘‘π‘Žπ‘›βˆ’1( 2 π‘₯βˆ’1

√2 )+𝐢

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

∫ 1βˆ’π‘₯+2π‘₯2βˆ’π‘₯3

π‘₯ (π‘₯2+1)2 |𝑑π‘₯=

πŸβˆ’ 𝒙+𝟐 π’™πŸβˆ’ π’™πŸ‘

𝒙 (π’™πŸ+𝟏)𝟐

𝑨(π’™πŸ+𝟏)𝟐+𝒙 ( 𝑩𝒙+π‘ͺ ) (π’™πŸ+𝟏)+𝒙 (𝑫𝒙+𝑬 )=βˆ’π’™πŸ‘+𝟐 π’™πŸβˆ’ 𝒙+𝟏

ΒΏ,

𝑨=𝟏 ,𝑩=βˆ’πŸ ,π‘ͺ=βˆ’πŸ ,𝑫=𝟏 ,𝑬=𝟎

¿∫ 1π‘₯βˆ’π‘₯+1

π‘₯2+1+

π‘₯(π‘₯2+1)2 𝑑π‘₯∫ 1βˆ’π‘₯+2π‘₯2βˆ’π‘₯3

π‘₯ (π‘₯2+1)2 𝑑π‘₯

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

¿∫ 1π‘₯𝑑π‘₯βˆ’βˆ« π‘₯

π‘₯2+1𝑑π‘₯βˆ’βˆ« 1

π‘₯2+1𝑑π‘₯+∫ π‘₯

(π‘₯2+1)2 𝑑π‘₯

∫ 𝒙(π’™πŸ+𝟏)𝟐

𝒅𝒙 , du = 2x dx

ΒΏπŸπŸβˆ«π’–βˆ’πŸπ’…π’–=

βˆ’πŸπŸ

π’–βˆ’πŸΒΏβˆ’πŸ

𝟐(π’™πŸ+𝟏)

∫ πŸπ’™πŸ+𝟏

𝒅𝒙=π’•π’‚π’βˆ’πŸπ’™

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

∫ √π‘₯+4π‘₯

𝑑π‘₯,

,

ΒΏ2∫ 𝑒2

𝑒2βˆ’4𝑑𝑒

¿2∫(1+4

𝑒2βˆ’4)𝑑𝑒

ΒΏ2∫1𝑑𝑒+8∫ 𝑑𝑒𝑒2βˆ’4

πŸπ’–πŸβˆ’πŸ’

=𝟏

(π’–βˆ’πŸ)(𝒖+𝟐)

¿𝑨

π’–βˆ’πŸ+

𝑩𝒖+𝟐

𝑨𝒖+𝟐 𝑨+π‘©π’–βˆ’πŸπ‘©=𝟏, ΒΏ2𝑒+2∫ 1

π‘’βˆ’2π‘‘π‘’βˆ’2∫ 1

𝑒+2𝑑𝑒

ΒΏ2√π‘₯+4+2 𝑙𝑛|√π‘₯+4βˆ’2|βˆ’2 𝑙𝑛|√π‘₯+4+2|+𝐢¿2√π‘₯+4+2 𝑙𝑛|√π‘₯+4βˆ’2

√π‘₯+4+2 |+𝐢