Question #62722

2 Answers
Jul 7, 2017

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

Jul 7, 2017

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