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