Question

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

Answer 1

`int3/((x-2)(x+1))dx=lnabs((x-2)/(x+1))+"c"`

Explanation:

In order to integrate the function, we first use partial fractions to split the integrand

`3/((x-2)(x+1))=3 1/((x-2)(x+1))=3 ((x+1)-(x-2))/(3(x-2)(x+1))=3 (((x+1))/(3(x-2)(x+1))-((x-2))/(3(x-2)(x+1)))=1/(x-2)-1/(x+1)`

Now integrating, we get

`3/((x-2)(x+1))dx=int1/(x-2)-1/(x+1)dx=lnabs(x-2)-lnabs(x+1)+"c"=lnabs((x-2)/(x+1))+"c"`

Answer 2

`int3/((x-2)(x+1))=ln|\x-2|+ln|\1/(x+1)|+C`

Explanation:

Partial Fractions

`3/((x-2)(x+1))=A/(x-2)+B/(x+1)`

Multiply the whole equation by `(x-2)(x+1)`

`3=A(x+1)+B(x-2)`

expand the brackets

`3=Ax+A+Bx-2B`

factorise

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

`A+B` has to equal 0 because there is no x term for the other side
and `A-2B` has to equal 3 because those terms have no x same as 3

So

`A+B=0` `(1)`
and
`A-2B=3` `(2)`

subtract equation `(2)` from equation `(1)` to get

`0-3B=3`
`(cancel(-3)B)/cancel(-3)=-cancel((3)/3)`

`:. B=-1`

Substitute `B` for 3 in either equation `(1)` or `(2)` to get `A`

`A+(-1)=0`
`:.A=1`

or

`A-2(-1)=3`
`:.A=1`

`:. 3/((x-2)(x+1))=1/(x-2)+(-1)/(x+1)`

Integration

`int3/((x-2)(x+1))dx=int1/(x-2)+(-1)/(x+1)dx`

`int1/(x-2)+(-1)/(x+1)dx=int1/(x-2)dx+int(-1)/(x+1)dx`

`int1/(x-2)dx+int(-1)/(x+1)dx=int1/(x-2)dx-int1/(x+1)dx`

because of the constant multiple of -1

`int1/(x-2)dx-int1/(x+1)dx`

Evaluate the two integrals separately

`int1/(x-2)dx`

use u-substitution

`u=x-2`
`(du)/dx=1`

`(cancel(dx)du)/cancel(dx)=1*dx`

Bring the integral into the u world

`int1/udu`

`ln|\u|+C_1`

Bring back into the x world

`ln|\x-2|+C_1`

second integral

`-int1/(x+1)dx`

use another letter substitution so you or the marker doesn't get confused

`v=x+1`

`(dv)/dx=1`

`(cancel(dx)dv)/cancel(dx)=1*dx`

bring the integral into the v world

`-int1/vdv`

`-(ln|\v|+C_2)`

bring back into the x world

`-(ln|\x+1|+C_2)`

A negative constant, adding or subtracting a constant from another constant is still a constant so we will just make another constant

`-ln|\x+1|+C_3`

Use the log law `alog_b(c)=log_b(c^a)`

`ln|\(x+1)^-1|+C_3`

`ln|\1/(x+1)|+C_3`

put the parts of the integral that we answered back together

`ln|\x-2|+ln|\1/(x+1)|+C_1+C_3`

`:.`

`ln|\x-2|+ln|\1/(x+1)|+C`