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∫3e2x+54e2x-5 dx - Mathematics and Statistics

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Question

`int (3"e"^(2x) + 5)/(4"e"^(2x) - 5)  "d"x`

Sum
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Solution

Let I = `int (3"e"^(2x) + 5)/(4"e"^(2x) - 5)  "d"x`

Let 3e2x + 5 = `"A"(4"e"^(2x) - 5) + "B" "d"/("d"x) (4"e"^(2x) - 5)`

= A(4e2x – 5) + B(8e2x)

∴ 3e2x + 5 = e2x(4A + 8B) − 5A

By equating the coefficients on both sides, we get

4A + 8B = 3 and −5A = 5

Solving these equations, we get

A = − 1 and B = `7/8`

∴ 3e2x + 5 = `-1(4"e"^(2x) - 5) + 7/8(8"e"^(2x))`

∴ I = `int (-1(4"e"^(2x) - 5) + 7/8(8"e"^(2x)))/(4"e"^(2x) - 5)  "d"x`

= `- int  "d"x + 7/8 int (8"e"^(2x))/(4"e"^(2x) - 5)  "d"x`

∴ I = `- x + 7/8  log|4"e"^(2x) - 5| + "c"`    .......`[∵ int  ("f'"(x))/("f"(x))  "d"x = log|"f"(x)| + "c"]`

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Chapter 2.3: Indefinite Integration - Long Answers III

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