Integrate:Root(1+root(x))/x
∫√(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
=
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))
To integrate the function ∫(√(1 + √x))/x, we can apply the technique of integration by substitution. Let's denote the expression inside the square root as u.
Let u = 1 + √x
Then, differentiating both sides with respect to x, we get:
du/dx = (1/2)*(1/√x) = (1/2√x)
Rearranging this equation, we find:
dx = 2√x * du
Substituting the values of x and dx, the integral becomes:
∫(√(1 + √x))/x * dx = ∫(√u)/(u - 1) * 2√x * du
Now we need to express x and dx in terms of u and du. To get x in terms of u, we square both sides of the equation u = 1 + √x.
u^2 = (1 + √x)^2 = 1 + 2√x + √x^2
√x^2 = u^2 - 2u + 1
Since we're interested in positive values of x, we take the positive root:
√x = √(u^2 - 2u + 1) = (u - 1)
Using this expression, dx becomes:
dx = 2√x * du = 2(u - 1) * du
Now, we can rewrite the integral in terms of u:
∫(√u)/(u - 1) * 2√x * du = 2∫(√u)/(u - 1) * (u - 1) * du = 2∫√u * du
= 2∫u^(1/2) * du
Integrating u^(1/2) with respect to u gives us:
2 * (2/3) * u^(3/2) + C = (4/3)u^(3/2) + C
Finally, substituting u back as 1 + √x, we have:
(4/3)(1 + √x)^(3/2) + C
Therefore, the integral of (√(1 + √x))/x is equal to (4/3)(1 + √x)^(3/2) + C, where C denotes the constant of integration.