# What is the integral of 1/((u-2)(u-1))? I don't know how to go about this question.

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You need to use partial fraction decomposition such that:

`1/((u-2)(u-1)) = A/(u-2) + B/(u-1)`

You need to bring the terms to the right to a common denominator such that:

`1 = A(u-1) + B(u-2)`

Opening the brackets yields:

`1 = A*u - A + B*u - 2B`

`1 = u(A+B) - A - 2B`

You need to equate the coefficients of like powers such that:

`{(A+B = 0),(-A-2B = 1):} => {(A=-B),(B-2B=1):}=>{(A=1),(B=-1):}`

`1/((u-2)(u-1)) = 1/(u-2)- 1/(u-1)`

Integrating both sides and using the property of linearity of integrals yields:

`int 1/((u-2)(u-1)) du= int 1/(u-2) du- int 1/(u-1) du`

`int 1/((u-2)(u-1)) du = ln |u-2| - ln|u-1| + c`

Using the logarithmic identities yields:

`int 1/((u-2)(u-1)) du = ln |(u-2)/(u-1)| + c`

**Hence, evaluating the given integral yields `int 1/((u-2)(u-1)) du = ln |(u-2)/(u-1)| + c.` **

Assume A and B are constant

1/ ((u-2)(u-1)) = A / (u-2) + B / (u-1)

Multiply both side by ((u-2)(u-1)),

1 = A(u-1) + B(u-2)

Assume u = 2,

A = 1

Assume u = 1,

B = -1

1/ ((u-2)(u-1)) = 1 / (u-2) - 1 / (u-1)

I = 1/ ((u-2)(u-1)) dx

=(1 / (u-2) - 1 / (u-1)) dx

= 1 / (u-2) dx - 1 / (u-1) dx

I = ln |u-2| - ln |u-1| + C