More Computing Integrals This particular problem has been giving me trouble, and while the math dept tutors did help a great deal, the resulting answer hasn't been accepted by the online homework submission website.  Find the definite integral of
$$\int\frac{x(x+1)}{2x^{3}+3x^{2}-13}$$
The work done so far with the help of the tutors is
$$\int\frac{x(x+1)}{2x^{3}+3x^{2}-13} = \int\frac{x^{2}+x}{2x^{3}+3x^{2}-13}$$
Let u=$2x^{3}+3x^{2}-13$
and ${u}'= 6x^{2}+6x = 6(x^{2}+x)$
$$\int\frac{1}{6}\cdot\frac{{u}'}{u}= \frac{1}{6}\int\frac{{u}'}{u}$$
$$\frac{1}{6}\ln~u= \frac{1}{6}\ln(2x^{3}+3x^{2}-13)+C$$
What is missing from this solution?
 A: What's missing is what you're integrating with respect to. You need to put a $ dx $ in the first 3 integrals of your post (and a $ du $ in any integrals involving $ u $).

Our integral is $ \int \frac{x^2 + x}{2x^3 + 3x^2 - 13} \ dx $. We make the substitution $ u = 2x^3 + 3x^2 - 13 $. This gives us $ \frac{du}{dx} = 6x^2 + 6x = 6(x^2 + x) $, and so $ dx = \frac{du}{6(x^2+x)} $. When we substitute this into the integral, the $ x^2 + x $ cancels to give $ \int \frac{1}{6} \frac{1}{u} \ du $. This evaluates to $ \frac{1}{6} \ln{|u|} + C $. We then substitute back in for $ x $, to leave $ \frac{1}{6}\ln{|2x^3 + 3x^2 - 13|} + C.$
Note that a substitution wasn't necessary. In general, $ \int \frac{f'(x)}{f(x)} \ dx = \ln{|f(x)|} + C $. In this case, we have something of the form $ \int \frac{ k f'(x)}{f(x)} \ dx $, where $ k $ is a constant. So we can integrate directly to give $ \int \frac{ k f'(x)}{f(x)} \ dx = k \ln{|f(x)|} + D $. You just need to spot that the numerator of the integrand is a constant multiple of the derivative of the denominator.
A: Possibly  the omission of the absolute value sign from 
$$\int dx/x = \ln|x| + C$$
could be the root of the problem.
A: ncmathsadist is probably right, it's the missing absolute value, but if I were writing the software I'd reject it for using $\ln$ instead of $\log$. I teach my classes that once you get to Calculus you have very little use for any logarithms other than natural ones, and the standard practice in higher math is to use $\log$ for the natural log. 
