Question

How do you evaluate the integral `int (x-2)/(3x(x+4))`?

Answer 1

`int(x-2)/(3x(x+4))dx=1/2ln(x+4)-1/6ln(x)+C`. See below.

Explanation:

The major component of the solution will be the use of partial fractions.

First, we can bring `1/3` outside the integral as a constant to simplify the integrand slightly.

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

Setting up partial fractions:

`(x-2)/(x(x+4))=A/x+B/(x+4)`

We multiply both sides by the denominator on the left, `x(x+4)`:

`[(x-2)/cancel(x(x+4))=A/cancelx+B/cancel(x+4)] (x(x+4))`

This gives:

`x-2=A(x+4)+Bx`

We pick values of `x` which will cancel one of the variables. If we try `x=0`, we'll get the `B` term to drop away (`B*0=0`).

`=>-2=A(4)`

Solving for `A`, we get `A=-1/2`.

Now we want to solve for B. If we set `x=-4` in the original equation, we'll get the `A` term to drop away:

`=>-6=B(-4)`

Solving for `B`, we get `B=3/2`. Now we put `A` and `B` back into our partial fractions:

`(x-2)/(x(x+4))=A/x+B/(x+4)`

`(x-2)/(x(x+4))=(-1/2)/x+(3/2)/(x+4)`

Substitute back into the integral:

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

We can break up the integral (sum rule):

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

Bringing the constants outside:

`-1/6int(1/x)dx+1/2int1/(x+4)dx`

For the first integral, we know that `int1/xdx=ln(x)`, so we have:

`-1/6ln(x)+1/2int1/(x+4)dx`

For the second integral, we can do a substitution, where `u=x+4` and `du=dx`. This gives:

`-1/6ln(x)+1/2int1/(u)du`

`=>-1/6ln(x)+1/2ln(u)`

Substitute back in for `u` and account for any constants:

`=>1/2ln(x+4)-1/6ln(x)+C`