# How do you integrate #int(x+1)/((x-3)(x-1)(x+4))# using partial fractions?

#2/7ln|x-3| - 1/5ln|x-1| - 3/35ln|x+4| + c#

since the factors on the denominator are linear , the numerators will be constants.

multiply through by (x-3)(x-1)(x+4)

x+ 1 = A(x-1)(x+4) + B(x-3)(x+4) + C(x-3)(x-1).................(1)

now require to find values of A , B and C. Note that if x=1 , the terms with A and C will be zero. If x =3 , the terms with B and C will be zero and if x = -4 , the terms with A and B will be zero. This is the starting point in finding values for A , B and C.

where c, is the constant of integration.

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To integrate ( \frac{{x + 1}}{{(x - 3)(x - 1)(x + 4)}} ) using partial fractions, first decompose it into partial fractions. The decomposition is as follows:

[ \frac{{x + 1}}{{(x - 3)(x - 1)(x + 4)}} = \frac{A}{{x - 3}} + \frac{B}{{x - 1}} + \frac{C}{{x + 4}} ]

To solve for ( A, B, ) and ( C ), multiply both sides of the equation by the denominator ( (x - 3)(x - 1)(x + 4) ) and then simplify.

[ x + 1 = A(x - 1)(x + 4) + B(x - 3)(x + 4) + C(x - 3)(x - 1) ]

Expanding and equating coefficients, you can solve for ( A, B, ) and ( C ). Once you have the values of ( A, B, ) and ( C ), integrate each term separately. This gives:

[ \int \frac{{x + 1}}{{(x - 3)(x - 1)(x + 4)}} , dx = \int \frac{A}{{x - 3}} , dx + \int \frac{B}{{x - 1}} , dx + \int \frac{C}{{x + 4}} , dx ]

[ = A \ln|x - 3| + B \ln|x - 1| + C \ln|x + 4| + D ]

Where ( D ) is the constant of integration.

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