Question

Question #62722

Answer 1

I tried this:

Explanation:

We can write:
`4int1/((x+3)(x-3))dx=`
and:
`=4int-1/6[1/(x+3)-1/(x-3)]dx=`
`=-4/6[int1/(x+3)dx-int1/(x-3)dx]=`
`=-2/3[ln|x+3|-ln|x-3|]+c`

Answer 2

`int4/(x^2-9)dx = 2/3[ln(abs(x-3))-ln(abs(x+3))]+C`

Explanation:

The first step to solving this integral is to break down the denominator into a factored form. We can recognize that the factored form of `x^2-9` is `(x-3)(x+3)`.

So rewriting the denominator we will get `int4/((x-3)(x+3))dx`.

Then we will take out the constant 4 to get `4int1/((x-3)(x+3))dx`.

We will then perform partial fraction decomposition. So setting aside the function itself, we will split it into two fractions:
`1/((x-3)(x+3)) = A/(x-3) + B/(x+3)`

Multiply our two new fractions by the old denominator and we will get: `1 = A(x+3) + B(x-3)`

We will solve for A by assuming x = 3 to get rid of B: `1 = A(6) + B(0)`, this is basically `1 = A(6)`.

Divide both sides by 6 and `A = 1/6`.

Now we need to solve for `B`. We will perform the same steps except this time we will set `x = -3` and we will get `1 = B(-6)`, which means `B = -1/6`.

Plug in our new fractions `1/(6(x-3)) - 1/(6(x+3))` into the integral to get: `4int1/(6(x-3)) - 1/(6(x+3))dx`.

Recognize that since both fractions share `1/6` we can take that out. So this will simplify to `(2/3)int1/((x-3)) - 1/((x+3))dx`

Now we will use substitution. Let's say `u = x-3` and `v = x+3`. The derivatives of these will simply be `du = dx` and `dv = dx`. So rewriting our integral:`(2/3)int1/(u)du - (2/3)int1/(v)dv`.

Remember that `int1/xdx = lnabsx+C`. So integrating our integrals we will get: `(2/3)(lnabsu - lnabsv)+C`.

Now we just plug back in u and v and our final answer will be:
`int4/(x^2-9)dx = 2/3[ln(abs(x-3))-ln(abs(x+3))]+C`