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∫3e2t+54e2t-5 dt - Mathematics and Statistics

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Question

`int (3"e"^(2"t") + 5)/(4"e"^(2"t") - 5)  "dt"`

Sum
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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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Chapter 1.5: Integration - Q.5

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