Solving Integral ∫(x^3)/(x^2 + 9) dx | Help with Integral Homework Statement

[Temp0]

Homework Statement

∫(x^3)/(x^2 + 9) dx

Homework Equations

The Attempt at a Solution

This question can be solved using long division, but I just wanted to know why I can't do it this other way.

So I start with one substitution, t = x^2, dt = 2xdx.
By taking out 1/2 from the integrand, I can make the integral:
∫(t dt)/(t + 9)
Then, using another substitution, u = t + 9, t = u - 9, du = dt
I make the equation into:
∫(u - 9)du / (u)

Separating the integrand into two separate integrals, I can solve it, and it becomes:
1/2 [(x^2 + 9) - 9ln(x^2 + 9)] + C
However, this isn't the right answer because the right answer does't contain a 9/2 constant inside. Why can't I solve this integral this way? Thanks for any help in advance.

 

[Mark44]

Homework Statement

∫(x^3)/(x^2 + 9) dx

Homework Equations

The Attempt at a Solution

This question can be solved using long division, but I just wanted to know why I can't do it this other way.

So I start with one substitution, t = x^2, dt = 2xdx.
By taking out 1/2 from the integrand, I can make the integral:
∫(t dt)/(t + 9)
Then, using another substitution, u = t + 9, t = u - 9, du = dt
I make the equation into:
∫(u - 9)du / (u)

Separating the integrand into two separate integrals, I can solve it, and it becomes:
1/2 [(x^2 + 9) - 9ln(x^2 + 9)] + C
However, this isn't the right answer because the right answer does't contain a 9/2 constant inside. Why can't I solve this integral this way? Thanks for any help in advance.

Your answer is correct, and differs from the result obtained by long division by a constant. After all, C and C + 9/2 are just constants. If you differentiate both answers, you get the integrand you started with.