integrate by parts
integrate (4+x^2)^1/2
I'd do a trig substitution first:
x = 2tanθ
4+x^2 = 4+tan^2θ = 4sec^2θ
dx = 2sec^2θ dθ
and the integral is now
∫2secθ * 2sec^2θ dθ
∫4sec^3θ dθ
now we can do the integration by parts. Let
u = secθ, du = secθtanθ dθ
dv = sec^2θ dθ, v = tanθ
∫u dv = uv - ∫v du
= 4secθtanθ - 4∫secθtan^2θ dθ
= 4secθtanθ - 4∫secθ(sec^2θ-1)dθ
= 4secθtanθ - 4∫sec^3θ dθ + 4∫secθ dθ
If we let I = ∫4sec^3θ dθ, we now have
I = 4secθtanθx - I + 4ln(secθ+tanθ)
2I = 4secθtanθ + 4ln(secθ+tanθ)
I = 2secθtanθ + 2ln(secθ+tanθ)
Now substitute back in for θ and we have
θ = arctan(x/2)
tanθ = x/2
secθ = 2/√(4+x^2)
∫√(4+x^2) dx = 2*√(4+x^2)/2*x/2 + 2ln(x/2 + √(4+x^2))
= x/2 √(4+x^2) + 2arcsinh(x/2)
oops. secθ = √(4+x^2)/2
To integrate the function (4+x^2)^(1/2) by parts, we can use the formula for integration by parts:
∫(u * v) dx = u * ∫v dx - ∫(u' * ∫v dx) dx
Let's assign u and dv to parts of the function:
u = (4+x^2)^(1/2) (function we differentiate)
dv = dx (function we integrate)
Next, we need to calculate du and v:
To calculate du (the derivative of u), we differentiate u with respect to x:
du/dx = d/dx[(4+x^2)^(1/2)] = 1/2(4+x^2)^(-1/2) * d/dx (4+x^2) = 1/2(4+x^2)^(-1/2) * 2x = x/(4+x^2)^(1/2)
To calculate v, we integrate dv (which is dx):
∫dx = x
Now, we can substitute u, v, du, and dv into the integration by parts formula:
∫(4+x^2)^(1/2) dx = u * v - ∫(u' * v) dx
∫(4+x^2)^(1/2) dx = (4+x^2)^(1/2) * x - ∫(x/(4+x^2)^(1/2) * x) dx
Now, we simplify the integral on the right-hand side:
∫(x/(4+x^2)^(1/2) * x) dx = x * ∫(x/(4+x^2)^(1/2)) dx
To evaluate this integral, we can make a substitution:
Let w = 4 + x^2
dw = 2x dx
x dx = (1/2) dw
Substituting the variables, we get:
∫(x/(4+x^2)^(1/2) * x) dx = (1/2) ∫(1/√w) dw = (1/2) * 2√w = √w = √(4 + x^2)
Plugging this back into the original equation, we get:
∫(4+x^2)^(1/2) dx = (4+x^2)^(1/2) * x - √(4 + x^2) + C
Therefore, the integral of (4+x^2)^(1/2) is given by:
∫(4+x^2)^(1/2) dx = (4+x^2)^(1/2) * x - √(4 + x^2) + C
where C is the constant of integration.