Question

निम्नलिखित के मान निकालिए-

`int  x/sqrt(x + 1)"d"x`  (संकेत: `sqrtx` = z रखिए)

Answer 1

I = `int x/(sqrt(x) + 1) "d"x` 

`sqrt(x)` = t रखिए

⇒ 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"`

अत:, I = `2[(xsqrt(x))/3 - x/2 + sqrt(x) - log |sqrt(x) + 1|] + "C"`