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प्रश्न
Evaluate the following:
`int x/(sqrt(x) + 1) "d"x` (Hint: Put `sqrt(x)` = z)
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उत्तर
I = `int x/(sqrt(x) + 1) "d"x`
Put `sqrt(x)` = t
⇒ x = t2
∴ dx = 2t . dt
∴ I = `int ("t" * 2"t" * "dt")/("t" + 1)`
= `2int "t"^3/("t" + 1) "dt"`
= `2int ("t"^3 + 1 - 1)/("t" + 1) "dt"`
= `2int ("t"^3 + 1)/("t" + 1) "dt" - 2int 1/("t" + 1) "dt"`
= `2int (("t" + 1)("t"^2 - "t" + 1))/("t" + 1) "dt" - 2int 1/("t" + 1) "dt"`
= `2int ("t"^2 - "t" + 1) "dt" - 2int 1/("t" + 1) "dt"`
= `2["t"^3/3 - "t"^2/2 + "t"] - 2 log |"t" + 1|`
= `2[x^(3/2)/3 - x/2 + sqrt(x)] - 2 log |sqrt(x) + 1| + "C"`
= `2[(xsqrt(x))/3 - x/2 + sqrt(x) - log |sqrt(x) + 1|] + "C"`
Hence, I = `2[(xsqrt(x))/3 - x/2 + sqrt(x) - log |sqrt(x) + 1|] + "C"`
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