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

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

We seek:

Then plugging into the IBP formula:

gives us

Then substituting intio the last integral, we get:

Restoring the substitution and using the result with [A] we get:

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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

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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