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Question
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
Solution
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = `"e"^x/(x + 4) + "C"`.
Explanation:
Let I = `int (x + 3)/(x + 4)^2 * "e"^x "d"x`
= `int (x + 4 - 1)/(x + 4)^2 * "e"^x "d"x`
= `int [(x + 4)/(x + 4)^2 - 1/(x + 4)^2]"e"^x "d"x`
= `int [1/(x + 4) - 1/(x + 4)^2]"e"^x "d"x`
Put `1/(x + 4)` = t
⇒ `- 1/(x + 4)^2 "d"x` = dt
Let f(x) = `1/(x + 4)`
∴ f'(x) = `- 1/(x + 4)^2`
Using `int "e"^x ["f"(x) + "f'"(x)]"d"x = "e"^x "f"(x) + "C"`
∴ I = `"e"^x * 1/(x + 4) + "C"`.
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