Integrate:Root(1+root(x))/x

Answer 1

∫√(1+√x) dx

Let u^2 = x
2u du = dx

Now you have

∫2u√(1+u) du

Now let v=1+u, dv = du

∫2(v-1)√v dv = 2∫v^(3/2) - v^(1/2) dv
= 2(2/5 v^(5/2) - 2/3 v^(3/2))
= 4/5 (1+u)^(5/2) - 4/3 (1+u)^(3/2)
= 4(1+u)^(3/2) ((1+u)/5 - 1/3)
= 4/15(1+√x)^(3/2) (3√x-2) + C
=

Answer 2

Man. I missed that x downstairs

∫√(1+√x)/x dx

Let u^2 = 1+√x
(u^2-1) = √
(u^2-1)^2 = x
2(u^2-1)(2u)du = dx

Now your integral is

∫4u^2/(u^2-1) du

as partial fractions, that is

∫4 + 4/(u-1) - 4/(u+1) du
= 4u + 4log((1-u)/(1+u))
= 4√(1+√x) + 4log((1-√(1+√x))/(1+√(1+√x)))
= 4√(1+√x) - 4tanh^-1(√(1+√x))