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प्रश्न
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
विकल्प
`1/(5x)(4 + 1/x^2)^-5 + "C"`
`1/5(4 + 1/x^2)^-5 + "C"`
`1/(10x)(1 + 4)^-5 + "C"`
`1/10(1/x^2 + 4)^-5 + "C"`
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उत्तर
`int x^9/(4x^2 + 1)^6 "d"x` is equal to `1/10(1/x^2 + 4)^-5 + "C"`.
Explanation:
Let I = `int x^9/(4x^2 + 1)^6 "d"x`
= `int x^9/(x^12(4 + 1/x^2)^6) "d"x`
= `int 1/(x^3(4 + 1/x^2)^6) "d"x`
Put `(4 + 1/x^2)` = t
⇒ `(-2)/x^3 "dt"` = dt
⇒ `"dx"/x^3 = - 1/2 "dt"`
∴ I = `- 1/2 int "dt"/"t"^6`
= `- 1/2 xx - 1/5 "t"^-5 + "C"`
= `1/10 "t"^-5 + "C"`
= `1/10(4 + 1/x^2)^-5 + "C"`
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