`int (5x-2) / (x-2)^2 dx` Use partial fractions to find the indefinite integral

`int(5x-2)/(x-2)^2dx`

Let's use partial fraction decomposition on the integrand,

`(5x-2)/(x-2)^2=A/(x-2)+B/(x-2)^2`

`5x-2=A(x-2)+B`

`5x-2=Ax-2A+B`

comparing the coefficients of the like terms,

`A=5`

`-2A+B=-2`

Plug in the value of A in the above equation,

`-2(5)+B=-2`

`-10+B=-2`

`B=-2+10`

`B=8`

So now `int(5x-2)/(x-2)^2dx=int(5/(x-2)+8/(x-2)^2)dx`

Now apply the sum rule,

`=int5/(x-2)dx+int8/(x-2)^2dx`

Take the constant's out,

`=5int1/(x-2)dx+8int1/(x-2)^2dx`

Now let's evaluate each of the above two integrals separately,

`int1/(x-2)dx`

Apply integral substitution `u=x-2`

`=>du=dx`

`=int1/udu`

Use the common integral :`int1/xdx=ln|x|`

`=ln|u|`

Substitute back `u=x-2`

`=ln|x-2|`

Now evaluate...

`int(5x-2)/(x-2)^2dx`

Let's use partial fraction decomposition on the integrand,

`(5x-2)/(x-2)^2=A/(x-2)+B/(x-2)^2`

`5x-2=A(x-2)+B`

`5x-2=Ax-2A+B`

comparing the coefficients of the like terms,

`A=5`

`-2A+B=-2`

Plug in the value of A in the above equation,

`-2(5)+B=-2`

`-10+B=-2`

`B=-2+10`

`B=8`

So now `int(5x-2)/(x-2)^2dx=int(5/(x-2)+8/(x-2)^2)dx`

Now apply the sum rule,

`=int5/(x-2)dx+int8/(x-2)^2dx`

Take the constant's out,

`=5int1/(x-2)dx+8int1/(x-2)^2dx`

Now let's evaluate each of the above two integrals separately,

`int1/(x-2)dx`

Apply integral substitution `u=x-2`

`=>du=dx`

`=int1/udu`

Use the common integral :`int1/xdx=ln|x|`

`=ln|u|`

Substitute back `u=x-2`

`=ln|x-2|`

Now evaluate the second integral,

`int1/(x-2)^2dx`

Apply integral substitution:`v=x-2`

`dv=dx`

`=int1/v^2dv`

`=intv^(-2)dv`

Apply the power rule,

`=v^(-2+1)/(-2+1)`

`=-v^(-1)`

`=-1/v`

Substitute back `v=x-2`

`=-1/(x-2)`

`:.int(5x-2)/(x-2)^2dx=5ln|x-2|+8(-1/(x-2))`

Add a constant C to the solution,

`=5ln|x-2|-8/(x-2)+C`

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