# How do you integrate #f(x)=x/(x^2+1)# using the quotient rule?

You throw the quotient rule away and write down

The quotient rule is from differentiation, not integration.

You can guess that the answer is something like

Notice how for large values of

In the graph above, the green line is the exact integral of the red line, and the blue line is the approximation achieved by ignoring the

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To integrate ( f(x) = \frac{x}{x^2 + 1} ) using the quotient rule, we need to apply the technique of integration by parts. First, let's express ( f(x) ) as the quotient of two functions ( u(x) ) and ( v(x) ):

[ u(x) = x ] [ v(x) = \frac{1}{x^2 + 1} ]

Now, we differentiate ( u(x) ) and ( v(x) ) to find ( du(x) ) and ( dv(x) ):

[ du(x) = dx ] [ dv(x) = \frac{-2x}{(x^2 + 1)^2} dx ]

Now, we apply the integration by parts formula:

[ \int u(x) , dv(x) = u(x)v(x) - \int v(x) , du(x) ]

Substitute the values of ( u(x) ), ( v(x) ), ( du(x) ), and ( dv(x) ) into the formula:

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

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

This new integral is a bit more complex. To solve it, you can use a trigonometric substitution or partial fraction decomposition followed by a trigonometric substitution. Once you integrate the remaining term, you'll have your final result.

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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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