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Integrate the following functions w.r.t. x : 4ex-252ex-5 - Mathematics and Statistics

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Question

Integrate the following functions w.r.t. x : `(4e^x - 25)/(2e^x - 5)`

Sum
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Solution

Let I = `int (4e^x - 25)/(2e^x - 5).dx`
Put,
Numerator = `"A (Denominator) + B"[d/dx ("Denominator")]`

∴ 4ex – 25 = `"A"(2e^x - 5) + "B"[d/dx(2e^x - 5)]`

= A(2ex – 5) + B(2ex – 0)

∴ 4ex – 25 = (2A + 2B)ex – 5A
Equating the coefficient of ex and constant on both sides, we get
2A + 2B = 4         ...(1)
and
5A = 25         
∴ A = 5
∴ from (1),2(5) + 2B = 4
∴ 2B = – 6
∴ B = – 3
∴ 4ex – 25 = 5(2ex –  5) –  3 (2ex)

∴ I = `int[(5(2e^xx - 5) - 3(2e^x))/(2e^x - 5)].dx`

= `int[5 - (3(2e^x))/(2e^x - 5)].dx`

= `5 int 1dx - 3 int (2e^x)/(2e^x - 5].dx`

= 5x – 3 log|2ex – 5| + c   ...`[∵ int (f'(x))/f(x)dx = log|f(x)| + c]` 

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Chapter 3: Indefinite Integration - Exercise 3.2 (A) [Page 110]

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