Page 42 - Integration Problems
P. 42
The integral becomes,
3 2 + 4
∫ 2
3 ( − 2)( + 4)
4 1 4
= ∫ − ∫ 2
3 ( − 2) 3 ( + 4)
4 1 1 4 2
= ∫ − ∫ 2
3 ( − 2) 2 3 ( + 4)
1
4
4
2
= [ln(| − 2|)] − [ln(| + 4|)]
3
3
2
= [ln(|4 − 2|) − ln(|3 − 2|)]
1
2
2
− [ln(|4 + 4|) − ln(|3 + 4|)]
2
1
= [ln(|2|) − ln(|1|)] − [ln(|20|) − ln(|13|)]
2
The keystrokes for this calculation (calculator in
COMP mode w1)
h2)p1a2$(h20)p
h13))=
38
