Aim: How do we integrate by partial fractions? (I)
Do Now: Write the partial fractions decomposition of
π΄π₯β1
+π΅π₯+2
=π΄ (π₯+2 )+π΅(π₯β 1)
(π₯β1)(π₯+2)π΄π₯+2π΄+π΅π₯βπ΅=π₯+5
( π΄+π΅ )π₯+(2 π΄βπ΅ )=π₯+5
π¨=π ,π©=βπ
β« π₯+5
π₯2+π₯β 2ππ₯
ΒΏβ«( 2π₯β1
β1
π₯+2 )ππ₯ΒΏ2 ππ|π₯β 1|βππ|π₯+2|+πΆ
π’=π₯β1 ,ππ’=ππ₯
π’=π₯+2,ππ’=ππ₯
A rational function whose numerator has higher degree than the denominator, divide and write the form of
β« π₯3+π₯π₯β 1
ππ₯ΒΏβ«(π₯2+π₯+2+2
π₯β1)ππ₯
ΒΏ π₯3
3+π₯
2
2+2π₯+2 ππ|π₯β 1|+πΆ
β« π₯2+2 π₯β12 π₯3+3π₯2 β2π₯
ππ₯
ΒΏβ« 12
1π₯
+15
12π₯β 1
+110
1π₯+2
ππ₯
ΒΏ12ππ|π₯|+ 1
10ππ|2 π₯β1|β 1
10ππ|π₯+2|+πΆ
In integrating the middle term we have made the mental substitution u = 2x β 1, which gives du = 2dx and
β« ππ₯π₯2βπ2
ΒΏ 12πβ«( 1
π₯βπβ
1π₯+π )ππ₯
π¨=πππ
,π©=βπππ
ΒΏ1
2π(ππ|π₯βπ|βππ|π₯+π|)+πΆ
ΒΏ 12π
ππ|π₯βππ₯+π|+πΆ
β« π₯4 β2π₯2+4 π₯+1π₯3βπ₯2 βπ₯+1
ππ₯ΒΏ π+π+π π
ππβππβ π+π
ππβπππ+π π+πππβ ππβπ+π
ΒΏβ« [π₯+1+ 1π₯β1
+ 2
(π₯β 1)2β
1π₯+1 ]ππ₯
ΒΏ π₯2
2+π₯+ ππ|π₯β1|β 2
π₯β1βππ|π₯+1|+πΆ
ΒΏ π₯2
2+π₯β
2π₯β1
+ππ|π₯β1π₯+1 |+πΆ
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