Answer 1

Let I = `int "x" sqrt(1 + "x"^2)` dx

Put 1 + x² = t

∴ 2x . dx = dt

∴ x . dx = `1/2` dt

∴ I = `1/2 int sqrt"t" * "dt"`

`= 1/2 int "t"^(1/2) *` dt

`= 1/2 * "t"^(3/2)/(3/2)` + c

`= 1/3 "t"^(3/2)` + c

∴ I = `1/3 (1 + "x"^2)^(3/2)` + c