# How do you evaluate the integral of #int x/[(x^2)+1]dx#?

Note that the derivative of the denominator is almost the numerator. Use substitution.

The integral is

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To evaluate the integral ∫(x/(x^2 + 1)) dx, you can use a trigonometric substitution. Let ( u = x^2 + 1 ), so ( du = 2x dx ). This substitution allows us to express the integral in terms of ( u ), which simplifies the integration.

Substitute ( u = x^2 + 1 ), which implies ( du = 2x dx ). Thus, ( dx = \frac{du}{2x} ).

Now, rewrite the integral in terms of ( u ):

[ \int \frac{x}{x^2 + 1} dx = \frac{1}{2} \int \frac{1}{u} du ]

Integrating ( \frac{1}{u} ) with respect to ( u ), we get:

[ \frac{1}{2} \ln|u| + C ]

Substitute back for ( u ):

[ \frac{1}{2} \ln|x^2 + 1| + C ]

So, the integral of ( \frac{x}{x^2 + 1} dx ) is ( \frac{1}{2} \ln|x^2 + 1| + C ), where ( C ) is the constant of integration.

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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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