# How do you evaluate the integral #int (lnx)^2dx#?

The answer is

We carry out integration piece by piece.

So,

We perform the partial integration once more.

so,

Combining everything at once

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To evaluate the integral ∫(ln(x))^2 dx, you can use integration by parts. Let u = (ln(x))^2 and dv = dx. Then, du = 2(ln(x))(1/x) dx and v = x. Applying the integration by parts formula:

∫u dv = uv - ∫v du

We have:

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

Simplify the integral:

∫(ln(x))^2 dx = x(ln(x))^2 - 2∫ln(x) dx

Now integrate ∫ln(x) dx separately:

∫ln(x) dx = xln(x) - ∫x * (1/x) dx = xln(x) - ∫dx

This simplifies to:

∫ln(x) dx = xln(x) - x + C

Where C is the constant of integration. Substitute this back into the previous expression:

∫(ln(x))^2 dx = x(ln(x))^2 - 2(xln(x) - x) + C

Finally, simplify:

∫(ln(x))^2 dx = x(ln(x))^2 - 2xln(x) + 2x + 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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