# How do you evaluate the integral #int ln(1+x^2)#?

Then:

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To evaluate the integral (\int \ln(1 + x^2) , dx), we can use integration by parts. Let (u = \ln(1 + x^2)) and (dv = dx), then (du = \frac{2x}{1 + x^2} , dx) and (v = x).

Applying the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Substitute the values:

[ \int \ln(1 + x^2) , dx = x \ln(1 + x^2) - \int \frac{2x^2}{1 + x^2} , dx ]

Now, the integral on the right-hand side is a rational function, which can be integrated using partial fraction decomposition or trigonometric substitution, depending on the preferred method.

After integrating the second term, you will have the final result for the integral (\int \ln(1 + x^2) , dx).

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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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