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Question
Evaluate the following.
`int (3"e"^"x" + 4)/(2"e"^"x" - 8)`dx
Solution
Let I = `int (3"e"^"x" + 4)/(2"e"^"x" - 8)`dx
Put, Numerator = A(Denominator) + B[`d/dx`(Denominator)]
Let 3ex + 4 = A(2ex - 8) + B `"d"/"dx"`(2ex - 8)
= 2 Aex - 8A + B(2ex )
∴ 3ex + 4 = (2A + 2B)ex - 8A
Comparing the coefficients of ex and constant term on both sides, we get
2A + 2B = 3 and - 8A = 4
Solving these equations, we get
A = `- 1/2` and B = 2
∴ I = `int (- 1/2 (2"e"^"x" - 8) + 2(2"e"^"x"))/(2"e"^"x" - 8)`dx
`= - 1/2 int "dx" + 2 int ("2e"^"x")/(2"e"^"x" - 8)` dx
∴ I = `- 1/2"x" + 2log |2"e"^"x" - 8|` + c ...`[int ("f" '("x"))/("f" ("x")) "dx" = log |f ("x")| + "c"]`
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