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?

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.
