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प्रश्न
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
विकल्प
`"e"^x/(1 + x^2) + "C"`
`(-"e"^x)/(1 + x^2) + "C"`
`"e"^x/(1 + x^2)^2 + "C"`
`(-"e"^x)/(1 + x^2)^2 + "C"`
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उत्तर
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to `"e"^x/(1 + x^2) + "C"`.
Explanation:
Let I = `int "e"^x ((1 - x)/(1 + x^2))^2 "d"x`
= `int "e"^x [(1 + x^2 - 2x)/(1 + x^2)^2]"d"x`
= `int "e"^x [((1 + x^2))/(1 + x^2)^2 - (2x)/(1 + x^2)^2]"d"x`
= `int "e"^x [1/(1 + x^2) - (2x)/(1 + x^2)^2]"d"x`
Here f(x) = `1/(1 + x^2)`
∴ f'(x) = `(-2x)/(1 + x^2)^2`
Using `int "e"^x ["f"(x) + "f'"(x)]"d"x = "e"^x * "f"(x) + "C"`
∴ I = `"e"^x * 1/(1 + x^2) + "C" = "e"^x/(1 + x^2) + "C"`
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