# What is #int ( 1+x)/(1+x^2)dx#?

The answer is

The integral is

Therefore,

And finally,

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To evaluate the integral (\int \frac{1+x}{1+x^2} , dx), you can use a simple substitution. Let (u = 1 + x^2), then (du = 2x , dx). Rearrange to solve for (dx), getting (dx = \frac{du}{2x}). Now substitute (u) and (dx) into the integral.

After substitution, the integral becomes (\int \frac{1}{u} , du), which integrates to (\ln|u| + C), where (C) is the constant of integration.

Finally, substitute back (u = 1 + x^2) to get the final result: (\ln|1 + x^2| + 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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