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प्रश्न
Integrate the following with respect to x :
`sqrt(x)/(1 + sqrt(x))`
उत्तर
`int sqrt(x)/(1 + sqrt(x)) "d"x`
Put `1 + sqrt(x)` = u
`1 + x^(1/2)` = u
`(0 + 1/2 x^(1/2 - 1)) "d"x` = du
`1/2 x^(- 1/2) "d"x` = du
`1/(2 x^(1/2)) "d"x` = du
`1/sqrt(x) "d"x` = 2 du
`int sqrt(x)/(1 + sqrt(x)) "d"x = int (sqrt(x) * sqrt(x))/(sqrt(x) * (1 + sqrt(x)) "d"x`
`int sqrt(x)/(1 + sqrt(x)) "d"x = int x/((1 + sqrt(x))) * 1/sqrt(x) "d"x` .......(1)
`1 + sqrt(x)` = u
`sqrt(x)= "u" - 1`
x = `("u" - 1)^2`
(1) ⇒ `int sqrt(x)/(1 + sqrt(x)) "d"x = int ("u" - 1)^2/"u" 2"du"`
= `2 int ("u"^2 - 2"u" + 1)/"u" "du"`
= `2 int ("u"^2/"u" - (2"u")/"u" + 1/"u") "du"`
= `2 int ("u" - 2 + 1/"u") "du"`
= `2[int "u" "du" - int 2 "du" + int 1/"u" "du"]`
= `2 ["u"^2/"u" - 2"u" + log|"u"|] + "c"`
= `2[(1 + sqrt(x))^2/2 - 2(1 + sqrt(x)) + log|1 + sqrt(x)|] + "c"`
`int sqrt(x)/(1 + sqrt(x)) "d"x = (1 + sqrt(x))^2 - 4(1 + sqrt(x)) + 2log|1 + sqrt(x)| + "c"`
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