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Question
`int (3"e"^(2"t") + 5)/(4"e"^(2"t") - 5) "dt"`
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Solution
Let I = `int (3"e"^(2"t") + 5)/(4"e"^(2"t") - 5) "dt"`
Let 3e2t + 5 = `"A"(4"e"^(2"t") - 5) + "B" "d"/"dt"(4"e"^(2"t") - 5)`
= 4Ae2t – 5A + B(8e2t)
∴ 3e2t + 5 = (4A + 8B) e2t – 5A
Comparing the coefficients of e2t and constant term on both sides,
we get 4A + 8B = 3 and – 5A = 5
Solving these equations,
we get A = – 1 and B = `7/8`
∴ I = `int (-1(4"e"^(2"t") - 5) + 7/8(8"e"^(2"t")))/(4"e"^(2"t") - 5) "dt"`
= `/int "dt" + 7/8 int (8"e"^(2"t"))/(4"e"^(2"t") - 5) "dt"`
∴ I = `-"t" + 7/8 log|4"e"^(2"t") - 5| + "c"` ......`[because int ("f'"(x))/("f"(x)) "d"x = log|"f"(x)| + "c"]`
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