Originally Posted by

**KevinShaughnessy** Hi,

I have an integral $\int \frac{6x^3}{x^3+1}$. First thing I do is long division to get $\int 6 - \int \frac{6}{x^3+1}$. To solve the second integral, I expand the denominator using the sum of cubes rule and then use partial fraction decomposition and find that A = 2, B = 2, and C = 4. This gives me the integrals $\int \frac{2}{x+1} + \int \frac{2x+4}{x^2-x+1}$. Solving the first one is easy, it's just $2 \ln (x+1)$. The second integral I think I solved correctly by first adding 5 and -5 to get $\int \frac{2x+4 - 5}{x^2-x+1} + \int \frac{5}{x^2-x+1}$. Now I can use u substitution on the first integral to get $\ln (x^2-x+1)$ and on the second integral I complete the square: $\int \frac{5}{(x-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}$. Solving this integral gives me $\frac{2}{\sqrt{3}}tan^{-1} \left( \frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)$.

Final answer is:

$2\ln (x+1) + \ln (x^2-x+1) + \frac{2}{\sqrt{3}}tan^{-1} \left( \frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) + 6x$

The problem is that when I enter this into my online assignment I'm told I'm wrong but I can't see my error. Does anyone see it?

Thanks.