Evaluate the integral # int (xlnx)/sqrt(x^2-1) #?

Question

Emily Ammerman · Accepted Answer

# int (xlnx)/sqrt(x^2-1) = lnxsqrt(x^2-1) - sqrt(x^2-1) + arctansqrt(x^2-1) + C #

We seek: # I = int (xlnx)/sqrt(x^2-1) # Let # { (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=x/sqrt(x^2-1), => v,=sqrt(x^2-1) ) :}# Then plugging into the IBP formula: # int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx # gives us # int \ (lnx)(x/sqrt(x^2-1)) \ dx = (lnx)(sqrt(x^2-1)) - int \ (sqrt(x^2-1))(1/x) \ dx # # :. I = lnxsqrt(x^2-1) - int \ sqrt(x^2-1)/x \ dx # ..... [A] For the next integral, #int \ sqrt(x^2-1)/x \ dx#, we can perform a substitution. Let #u=sqrt(x^2-1) => (du)/dx=x/sqrt(x^2-1)# Then substituting intio the last integral, we get: # int \ sqrt(x^2-1)/x \ dx = int \ u^2/(u^2+1) \ du# # " " = int \ (u^2+1-1)/(u^2+1) \ du# # " " = int \ 1 - 1/(u^2+1) \ du# # " " = u - arctanu# Restoring the substitution and using the result with [A] we get: # I = lnxsqrt(x^2-1) - {sqrt(x^2-1) - arctansqrt(x^2-1) } + C# # \ \ = lnxsqrt(x^2-1) - sqrt(x^2-1) + arctansqrt(x^2-1) + C #

Charles Anctil · Answer

undefined To evaluate the integral ∫(xlnx)/√(x^2-1), we use integration by parts. Let u = ln(x) and dv = x/√(x^2-1)dx. Then differentiate u to get du = (1/x)dx and integrate dv to get v = √(x^2-1). Now, we can apply the integration by parts formula:

∫u dv = uv - ∫v du

Substituting our values:

∫(xlnx)/√(x^2-1)dx = ln(x)√(x^2-1) - ∫√(x^2-1) * (1/x)dx

The integral on the right can be simplified using a trigonometric substitution. Let x = sec(θ), then dx = sec(θ)tan(θ)dθ. Substituting these values:

∫√(x^2-1) * (1/x)dx = ∫tan(θ)dθ

Integrating this gives:

= -ln|sec(θ)| + C

Now, recall that x = sec(θ), so sec(θ) = x. Therefore:

= -ln|x| + C

Thus, the final result of the integral is:

∫(xlnx)/√(x^2-1)dx = ln(x)√(x^2-1) + ln|x| + C

Evelyn Agard · Answer

undefined To evaluate the integral ∫(xlnx)/√(x^2-1), you can use the substitution method. Let u = x^2 - 1. Then, du/dx = 2x, and dx = du/(2x). Substituting these into the integral, we get:

∫(xlnx)/√(x^2-1) dx = ∫(xlnx)/(√u) * (du/(2x)).

Simplify to get:

(1/2)∫lnx * (du/√u).

Now, we can integrate this by applying the u-substitution rule for integrals, where we let v = ln(x), and dv = (1/x)dx:

(1/2)∫v * dv.

Integrating v * dv gives (1/2)(v^2/2) + C.

Substituting back v = ln(x), we get (1/4)(ln^2(x)) + C.

So, the integral of (xlnx)/√(x^2-1) with respect to x is (1/4)(ln^2(x)) + C.