Question

How do you integrate `int (4x+3) / (x - 1)^2dx` using partial fractions?

Answer 1

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Explanation:

`int (4(x-1)+4+3)/(x-1)^2 dx`

By substituting `x` in `4x` with the denominator, the fraction is split into partial fractions. Include a `+4` so the original fraction isn't changed.

`int (4cancel(x-1))/(x-1)^cancel2 +7/(x-1)^2dx`

Split the integrals to make it easier.

`=int 4/(x-1) dx + int 7/(x-1)^2 dx`

`=4 int 1/(x-1)dx + 7int(x-1)^-2dx`

1. `4 int 1/(x-1)dx = 4ln|x-1| + c_1` Integration of inverse functions.

2. `7 int(x-1)^-2dx = 7*-1*(x-1)^-1 + c_2` Using reverse chain rule.

Therefore,

`=4 int 1/(x-1)dx + 7int(x-1)^-2dx`

`=4ln|x-1| - 7(x-1)^-1 +c_3`