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.

  • $\begingroup$ Does this help? $\endgroup$ – azif00 Jan 28 at 20:49
  • $\begingroup$ Writing $x^2 + x + 3 = x(x+1) + 3$ may help slightly. $\endgroup$ – player3236 Jan 28 at 20:50
  • $\begingroup$ You can expand the numerator and use Horner's method for division. It's a bit faster. $\endgroup$ – Bernard Jan 28 at 20:50
  • $\begingroup$ @player3236 Especially if we also write $x^3+7$ as $x^3+1+6$, so the integrand is $(x(x+1)+3)(x^2-x+1+6/(x+1))$. $\endgroup$ – J.G. Jan 28 at 21:28

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.

  • 1
    $\begingroup$ This is my favorite approach to partial fractions with a linear denominator. $\endgroup$ – Chickenmancer Jan 29 at 2:19
  • $\begingroup$ Wow this is so much less time consuming than my approach. Thanks! $\endgroup$ – Ski Mask Jan 29 at 9:01

Horner's method for division:


\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 $\;18$.

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    $\begingroup$ +1 Synthetic division blew my mind when I was first taught it, and still continues to do so. A great tip not enough students know about. $\endgroup$ – Deepak Jan 29 at 1:46
  • $\begingroup$ Isn't it taught in high-school? $\endgroup$ – Bernard Jan 29 at 1:49
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    $\begingroup$ During my time (about 1989) in Singapore, I learned it in Secondary school (when I was 14 going on 15). It wasn't a method featured in the (UK-sourced) textbooks of the time (which only covered the longhand method of polynomial division) but the teacher taught this as a useful shortcut. I'm not sure if it was universally taught in all schools at that level (the school I attended has generally been regarded as the best in the country) and I don't know the situation now. $\endgroup$ – Deepak Jan 29 at 1:57
  • $\begingroup$ +1 for using this technique. I don't think it's a part of typical high school curriculum, but it is not something difficult and very useful. $\endgroup$ – Paramanand Singh Jan 29 at 2:39
  • $\begingroup$ @Bernard Wow thanks for the answer! I know that you can use Horner's Method to convert between binary and decimal numbers but this is the first time I have seen it used to divide polynomials. Nice answer. $\endgroup$ – Ski Mask Jan 29 at 9:02

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.

  • $\begingroup$ Haha saving ibuprofen. +1 good advice one may need the tablet in future for difficult problems. $\endgroup$ – Paramanand Singh Jan 29 at 2:36

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