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Question
Integrate the following functions w.r.t. x : `(3e^(2x) + 5)/(4e^(2x) - 5)`
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Solution
Let I = `int (3e^(2x) + 5)/(4e^(2x) - 5).dx`
Put,
Numerator = `"A (Denominator) + B"[d/dx("Denominator")]`
∴ 3e2x + 5 = `"A"(4e^(2x) - 5) + "B"[d/dx(4e^(2x) - 5)]`
= A(4e2x – 5) + B(4.e2x x 2 – 0)
∴ 3e2x + 5 = (4A + 8B)e2x – 5A
Equating the coeffiecient of e2x and constant on both sides, we get
4A + 8B = 3 ...(1)
and
– 5A = 5
∴ A = – 1
∴ from (1), 4(– 1) + 8B = 3
∴ 8B = 7
∴ B = `(7)/(8)`
∴ 3e2x + 5 = `- (4e^(2x) - 5) + 7/8(8e^(2x))`
∴ I = `int[(-(4e^(2x) - 5) +7/8(8e^(2x)))/(4e^(2x) - 5)].dx`
= `int[-1 +(7/8(8e^(2x)))/(4e^(2x) - 5)].dx`
= `int 1 dx + (7)/(8) int (8e^(2x))/(4e^(2x) - 5).dx`
= `- x + (7)/(8)log|4e^(2x) - 5| + c ...[∵ int (f'(x))/f(x)dx = log|f(x)| + c]`
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