Question

integrate 1/(x^5*(x^7)-1)
x^5*((x^7) - 1) is in denominator ​

Answer 1

Answer:

1/(x-5)(x-7)

= A/(x-5) + B/(x-7)

= [A(x-7) + B(x-5)]/(x-5)(x-7)

= [Ax - 7A + Bx - 5B]/(x-5)(x-7)

= [x(A+B) - (7A+5B)]/(x-5)(x-7)

Equate the numerator on both sides to get

1 = x(A+B) - (7A+5B). Form we get

A + B = 0 or

7A + 7B = 0 …(1)

7A + 5B = -1 …(2)

Subtract (2) from (1),

7B - 5B = 1, or

2B = -1, or B = -1/2 and so A = 1/2.

So the partial fractions are

= 1/2(x-5) - 1/2(x-7). Answer

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Answer 2

$\huge\red{Answer}$

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☆☞ [ Verified answer ]☜☆

1/(x-5)(x-7)

= A/(x-5) + B/(x-7)

= [A(x-7) + B(x-5)]/(x-5)(x-7)

= [Ax - 7A + Bx - 5B]/(x-5)(x-7)

= [x(A+B) - (7A+5B)]/(x-5)(x-7)

Equate the numerator on both sides to get

1 = x(A+B) - (7A+5B). Form we get

A + B = 0 or

7A + 7B = 0 …(1)

7A + 5B = -1 …(2)

Subtract (2) from (1),

7B - 5B = 1, or

2B = -1, or B = -1/2 and so A = 1/2.

So the partial fractions are

= 1/2(x-5) - 1/2(x-7). Answer

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