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How do you find #int x/(x^2-x-2)dx# using partial fractions?

Answer 1

Emilia Aguilar

#int x/(x^2-x-2) dx = 2/3 ln abs (x-2)+1/3 ln abs(x+1) + C#

#x/(x^2-x-2) = x/((x-2)(x+1)) = 2/(3(x-2))+1/(3(x+1))#

So:

#int x/(x^2-x-2) dx = int 2/(3(x-2))+1/(3(x+1)) dx#

#color(white)(int x/(x^2-x-2) dx) = 2/3 ln abs (x-2)+1/3 ln abs(x+1) + C#

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Answer 2

Cameron Alexander

To find ( \int \frac{x}{x^2 - x - 2} , dx ) using partial fractions, follow these steps:

Factor the denominator ( x^2 - x - 2 ) to obtain its roots.
Write the fraction ( \frac{x}{x^2 - x - 2} ) as the sum of two fractions with undetermined coefficients, ( \frac{A}{x - r_1} + \frac{B}{x - r_2} ), where ( r_1 ) and ( r_2 ) are the roots of the denominator.
Clear the fractions by finding a common denominator.
Equate the numerators of the original fraction and the partial fraction decomposition.
Solve the resulting system of equations for the coefficients ( A ) and ( B ).
Integrate each term separately.
Combine the results to obtain the final integral expression.

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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.