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

Answer 1

You throw the quotient rule away and write down #1/2 ln(x^2+1)#

The quotient rule is from differentiation, not integration.
You can guess that the answer is something like #ln(...)# because the differentiation #ln(...)# gives something like #1/(...)#. You then spot that the top is (give or take a constant factor) what you get by differentiating the bottom, #(x^2+1)# yielding #2x#. So try differentiating #ln(x^2+1)# using the chain rule and compare what you get with the question. In essence, you have to spot the `reverse chain rule'.

Notice how for large values of #x#, both positive and negative, the answer gets close to #log|x|#. This is because the #1# becomes negligible compared to the #x^2#, so the answer gets close to #(1/2)ln(x^2)#, which is #ln |x|#.

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 #1#, for suitably chosen values of the constant of integration. If you imagine these graphs being repeated with successively smaller values instead of #1# (say #0.1#, #0.01#, ... ) then the red graph sort of approaches the curve #y=1/x# (which has a vertical asymptote #x=0#), but "panics" at the thought of infinity and takes a short cut through the origin to the corresponding point on the other side!

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Answer 2

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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Answer from HIX Tutor

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.

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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