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

Answer 1

Do you mean Differentiate?

To integrate you might be able to use partial fraction but differentiation using the quotient rule will give you:

#(x'(x^2+1)-(x^2+1)'(x))/(x^2+1)^2#=#(x^2+1-3x)/(x^2+1)^2#
#y'#=#(x^2-3x+1)/(x^4+2x^2+1)#
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Answer 2

To integrate ( y = \frac{x}{x^2 + 1} ) using the quotient rule, follow these steps:

  1. Differentiate the numerator and denominator separately.
  2. Apply the quotient rule: (\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}).
  3. Integrate the resulting expression.

The quotient rule states: [ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} ]

Differentiate the numerator (x) and denominator (x^2 + 1) separately to find their derivatives:

[f'(x) = 1 \text{ and } g'(x) = 2x]

Apply the quotient rule:

[ \frac{d}{dx}\left(\frac{x}{x^2 + 1}\right) = \frac{(1)(x^2 + 1) - (x)(2x)}{(x^2 + 1)^2} ]

Simplify the expression:

[ \frac{d}{dx}\left(\frac{x}{x^2 + 1}\right) = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2} ]

Integrate the resulting expression:

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

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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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