How do you use partial fractions to find the integral #int (x^3-x+3)/(x^2+x-2)dx#?
THe answer is
The denominator is
But before, let's do a long division
So,
Now we do the decomposition in partial fractions
so,
So,
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To find the integral ( \int \frac{{x^3 - x + 3}}{{x^2 + x - 2}} , dx ) using partial fractions, first factor the denominator as ( (x - 1)(x + 2) ). Then, express the fraction as a sum of partial fractions:
[ \frac{{x^3 - x + 3}}{{x^2 + x - 2}} = \frac{{A}}{{x - 1}} + \frac{{B}}{{x + 2}} ]
Next, multiply both sides by the denominator ( (x - 1)(x + 2) ) to clear the fractions. This gives:
[ x^3 - x + 3 = A(x + 2) + B(x - 1) ]
Now, solve for ( A ) and ( B ) by equating coefficients of corresponding powers of ( x ). Once you find ( A ) and ( B ), integrate each term separately to find the integral.
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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.
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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