Evaluate $\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$. 
Evaluate:
$$\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$$

The only thing I can think of doing here is long division to simplify the integral down and see if I can work with some easier sections. Here's my attempt:
\begin{align}
\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx &= \int \left(\:x^4+3x^2+4x+3+\frac{18}{x+1} \right )dx \\
&= \int \:x^4dx+\int \:3x^2dx+\int \:4xdx+\int \:3dx+\int \frac{18}{x+1}dx \\
&= \frac{x^5}{5}+x^3+2x^2+3x+18\ln \left|x+1\right|+ c, c \in \mathbb{R}
\end{align}
The only issue I had is that the polynomial long division took quite some time. Is there another way to do this that is less time consuming? The reason I ask this is that, this kind of question can come in an exam where time is of the essence so anything that I can do to speed up the process will benefit me greatly.

As substitution by parts goes nowhere It simply do what you did but skip the step of multiplying out the polynomial $ \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}=\frac {(x(x+1) + 3)(x^3+7)}{x+1}= x(x^3 +7) + \frac 3{x+1}(x^3 + 7)=$
$x(x^3 +7) +\frac 3{x+1}(x^2(x+1) -x(x+1) + x+1 + 6)=x(x^3 +7) + 3(x^2-x+1) + \frac {18}{x+1} $
Saves one or two tablets of ibuprofen.
