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.
 A: Horner's method for division:
$$(x^2+x+3)(x^3+7)=x^5+x^4+3x^3+7x^2+7x+21$$
\begin{array}{*{7}{r}}
& 1 & 1  & 3 & 7 & 7 & 21 \\ 
+ & \downarrow & -1 & 0 & -3 & -4 & -3 \\
\hline
\times -1 & \color{red}1 &  \color{red}0 &  \color{red}3 &  \color{red}4 &  \color{red}3 & \color{cyan}{18}
\end{array}
so the quotient is $\;x^4+3x^2+4x+3$ and the remainder is $\;18$.
A: 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.
A: If you do the substitution $x+1=t\iff x=t-1$ you only need do a (long) multiplication:
$$\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx=
\int \frac{\left((t-1)^2+t+2\right)\left((t-1)^3+7\right)}{t}dt=
$$
$$=\int \frac{\left(t^2-t+3\right)\left(t^3-3t^2+3t+6\right)}{t}dt=\int \frac{t^5-4 t^4+9 t^3-6 t^2+3 t+18}{t}dt=$$
$$=\int\left( t^4-4 t^3+9 t^2-6 t+3 +\frac{18}{t}\right)dt
$$
And now it's simple.
