How do you integrate #f(x)=(x-2)/((x^2+5)(x-3)(x-1))# using partial fractions?
The answer is
Let's perform the decomposition into partial fractions
Therefore,
Therefore,
So,
So,
Putting it all together
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To integrate the function ( f(x) = \frac{x - 2}{(x^2 + 5)(x - 3)(x - 1)} ) using partial fractions, follow these steps:
- Factor the denominator ( (x^2 + 5)(x - 3)(x - 1) ).
- Decompose the fraction into partial fractions.
- Solve for the constants.
- Integrate each partial fraction separately.
- Combine the integrals to find the overall integral of ( f(x) ).
Step 1: Factor the denominator: [ (x^2 + 5)(x - 3)(x - 1) = (x^2 + 5)(x^2 - 4x + 3) ]
Step 2: Decompose into partial fractions: [ \frac{x - 2}{(x^2 + 5)(x - 3)(x - 1)} = \frac{A}{x^2 + 5} + \frac{B}{x - 3} + \frac{C}{x - 1} ]
Step 3: Solve for the constants A, B, and C. [ x - 2 = A(x - 3)(x - 1) + B(x^2 + 5)(x - 1) + C(x^2 + 5)(x - 3) ]
Step 4: Solve for A, B, and C by substituting appropriate values of x. [ A = \frac{3}{32}, \quad B = -\frac{1}{8}, \quad C = \frac{1}{32} ]
Step 5: Integrate each partial fraction separately: [ \int \frac{A}{x^2 + 5} , dx = \frac{A}{\sqrt{5}} \arctan{\left(\frac{x}{\sqrt{5}}\right)} + C_1 ] [ \int \frac{B}{x - 3} , dx = B \ln{|x - 3|} + C_2 ] [ \int \frac{C}{x - 1} , dx = C \ln{|x - 1|} + C_3 ]
Step 6: Combine the integrals and constants: [ \int \frac{x - 2}{(x^2 + 5)(x - 3)(x - 1)} , dx = \frac{3}{32\sqrt{5}} \arctan{\left(\frac{x}{\sqrt{5}}\right)} - \frac{1}{8} \ln{|x - 3|} + \frac{1}{32} \ln{|x - 1|} + C ]
Where ( C ) is the constant of integration.
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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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