Please answer ASAP >.< Question about Integration

I get the following for (a):

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a)

Take the integral:
 integral1/(x^2 - 9) dx
Factor -9 from the denominator:
 = integral-1/(9 (1 - x^2/9)) dx
Factor out constants:
 = -1/9 integral1/(1 - x^2/9) dx
For the integrand 1/(1 - x^2/9), substitute u = x/3 and du = 1/3 dx:
 = -1/3 integral1/(1 - u^2) du
The integral of 1/(1 - u^2) is tanh^(-1)(u):
 = -1/3 tanh^(-1)(u) + constant
Substitute back for u = x/3:
 = -1/3 tanh^(-1)(x/3) + constant
Which is equivalent for restricted x values to:
Answer: |= 1/6 (log(3 - x) - log(x + 3)) + constant

b)

Take the integral:
 integral1/((x + 3) (x^2 + 9)) dx
For the integrand 1/((x + 3) (x^2 + 9)), use partial fractions:
 = integral((3 - x)/(18 (x^2 + 9)) + 1/(18 (x + 3))) dx
Integrate the sum term by term and factor out constants:
 = 1/18 integral(3 - x)/(x^2 + 9) dx + 1/18 integral1/(x + 3) dx
Expanding the integrand (3 - x)/(x^2 + 9) gives 3/(x^2 + 9) - x/(x^2 + 9):
 = 1/18 integral(3/(x^2 + 9) - x/(x^2 + 9)) dx + 1/18 integral1/(x + 3) dx
Integrate the sum term by term and factor out constants:
 = -1/18 integral x/(x^2 + 9) dx + 1/6 integral1/(x^2 + 9) dx + 1/18 integral1/(x + 3) dx
For the integrand x/(x^2 + 9), substitute u = x^2 + 9 and du = 2 x dx:
 = -1/36 integral1/u du + 1/6 integral1/(x^2 + 9) dx + 1/18 integral1/(x + 3) dx
The integral of 1/u is log(u):
 = -(log(u))/36 + 1/6 integral1/(x^2 + 9) dx + 1/18 integral1/(x + 3) dx
Factor 9 from the denominator:
 = -(log(u))/36 + 1/6 integral1/(9 (x^2/9 + 1)) dx + 1/18 integral1/(x + 3) dx
Factor out constants:
 = -(log(u))/36 + 1/54 integral1/(x^2/9 + 1) dx + 1/18 integral1/(x + 3) dx
For the integrand 1/(x^2/9 + 1), substitute s = x/3 and ds = 1/3 dx:
 = -(log(u))/36 + 1/18 integral1/(s^2 + 1) ds + 1/18 integral1/(x + 3) dx
The integral of 1/(s^2 + 1) is tan^(-1)(s):
 = 1/18 tan^(-1)(s) - (log(u))/36 + 1/18 integral1/(x + 3) dx
For the integrand 1/(x + 3), substitute p = x + 3 and dp = dx:
 = 1/18 tan^(-1)(s) - (log(u))/36 + 1/18 integral1/p dp
The integral of 1/p is log(p):
 = (log(p))/18 + 1/18 tan^(-1)(s) - (log(u))/36 + constant
Substitute back for p = x + 3:
 = 1/18 tan^(-1)(s) - (log(u))/36 + 1/18 log(x + 3) + constant
Substitute back for s = x/3:
 = -(log(u))/36 + 1/18 log(x + 3) + 1/18 tan^(-1)(x/3) + constant
Substitute back for u = x^2 + 9:
 = -1/36 log(x^2 + 9) + 1/18 log(x + 3) + 1/18 tan^(-1)(x/3) + constant
Which is equal to:
Answer: |= 1/36 (-log(x^2 + 9) + 2 log(x + 3) + 2 tan^(-1)(x/3)) + constant

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