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प्रश्न
`int 1/(x^2 + 1)^2`dx = __________.
विकल्प
`tan^-1 x - 1/(2x(x^2 + 1)) + "c"`
`1/2 tan^-1 x + x/(2(x^2 + 1)) + "c"`
`tan^-1 x + 1/(x^2 + 1) + "c"`
`tan^-1 x + 1/(2(x^2 + 1)) + "c"`
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उत्तर
`int 1/(x^2 + 1)^2`dx = `underline(1/2 tan^-1 x + x/(2(x^2 + 1)) + "c")`.
Explanation:
`int 1/(x^2 + 1)^2 "dx" = 1/(x^2 + 1)`
`int 1 "dx" - int ("d"/"dx" (1/(x^2 + 1)) int 1 "dx") "dx"` ...(Integration by parts)
`= x/(x^2 + 1) - int (- 2x)/(x^2 + 1)^2 x "dx"`
`= x/(x^2 + 1) + 2 int (x^2 + 1 - 1)/(x^2 + 1)^2`dx
`= x/(x^2 + 1) + 2 int x/(x^2 + 1) "dx" - 2 int "dx"/(x^2 + 1)^2` ....(i)
And also, we know that
`int "dx"/(x^2 + 1) = tan^-1 (x)` ...(ii)
From Eqs. (i) and (ii), we get
`tan^-1 x = x/(x^2 + 1) + 2tan^-1 x - 2 int "dx"/(x^2 + 1)^2`
`=> 2 int "dx"/(x^2 + 1)^2 = tan^-1 x + x/(x^2 + 1)`
`=> int "dx"/(x^2 + 1)^2 = 1/2 tan^-1 x + x/(2((x^2 + 1)))`+ c
