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प्रश्न
Evaluate the following.
`int 1/(sqrt"x" + "x")` dx
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उत्तर
Let I = `int 1/(sqrt"x" + "x")` dx
= `int 1/(sqrt"x" (1 + sqrt"x"))`dx
Put `1 + sqrt"x" = "t"`
∴ `1/(2sqrt"x") "dx" = "dt"`
∴ `1/sqrt"x"`dx = 2 dt
∴ I = `int (2 * "dt")/"t"`
`= 2 int 1/"t" * "dt"`
= 2 log | t | + c
∴ I = 2 log `|1 + sqrt"x"|` + c
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