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Question
`int tan^-1 sqrt(x) "d"x` is equal to ______.
Options
`(x + 1) tan^-1 sqrt(x) - sqrt(x) + "C"`
`x tan^-1 sqrt(x) - sqrt(x) + "C"`
`sqrt(x) - x tan^-1 sqrt(x) + "C"`
`sqrt(x) - (x + 1) tan^-1 sqrt(x) + "C"`
Solution
`int tan^-1 sqrt(x) "d"x` is equal to `(x + 1) tan^-1 sqrt(x) - sqrt(x) + "C"`.
Explanation:
Let I = `int 1 * tan^-1 sqrt(x) "d"x`
= `tan^-1 sqrt(x) int 1 "d"x - int[(tan^-1 sqrt(x))"'" int 1"d"x]"d"x`
= `tan^-1 sqrt(x) * x - int 1/(1 + x) * 1/(2sqrt(x)) * x"d"x` ....[Integrating by parrts]
= `xtan^-1 sqrt(x) - 1/2 int sqrt(x)/(1 + x) "d"x`
Put x = t2
⇒ dx = 2t dt
∴ I = `xtan^-1 sqrt(x) - int "t"^2/(1 + "t"^2) "d"x`
= `xtan^-1 sqrt(x) - int (1 - 1/(1 + "t"^2))"dt"`
= `xtan^-1 sqrt(x) - "t" + tan^-1 1 + "C"`
= `xtan^-1 sqrt(x) - sqrt(x) + tan^-1 sqrt(x) + "C"`
= `(x + 1) tan^-1 sqrt(x) - sqrt(x) + "C"`
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