Question

Evaluate the following.

`int 1/(sqrt"x" + "x")` dx

Answer 1

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