I've been trying this integral

$$\int \frac{x^2+x}{(e^x+x+1)^2}dx$$

For quite some time now but I am stuck on it. The things I tried include factoring numerator as $x(x+1)$ and expanding denominator as $$\big( e^x +(x+1) \big)^2$$ but I'm unable to solve it. I found a solution online which used integration by parts however this question can supposedly be solved by substitution according to my teacher and does not involve using partial fractions as well.

  • $\begingroup$ I found a perfectly fine method which is using only substitution. Should I share it? (incase anyone want it) $\endgroup$ – user671231 Jul 16 '19 at 15:20

The following development may seem unmotivated, but it does work. Let $$ v := e^x, \;\; w := 1 + x + v, \;\; w' = 1 + v = w - x. $$ We are trying to integrate $$ u := \frac{x^2+x}{(e^x+x+1)^2} $$ Notice the equalities $$ u = \frac{x (1 + x)}{w^2} = \frac{x (w - v)}{w^2} = \frac{x}{w} - \frac{x v}{w^2} = 1 - \frac{w - x}w - \frac{x v}{w^2} = 1 - \frac{w'}w - \frac{ x v}{w^2}$$ We suppose for some unknown $\,t\,$ $$ y := \frac{t}{w} \;\; \text{ and } \;\; y' = \frac{t'\, w - t\, w'}{w^2} = \frac{t'\, w - t\, (w-x)}{w^2} = \frac{(t'-t)w + t x}{w^2} = -\frac {x v}{w^2}. $$ To solve this last equation, notice that $$ (t' - t)(1 + x + v) + t x = - x v $$ solved by algebra implies that $\,t = 1+x\,$ and $\, t' = 1.\,$ Now $$ \int u\,dx = C + x - \ln(w) + \frac{1+x}{w}. $$

In a situation like this, it helps to have lots of practice with integrals ranging from simple to more complicated ones. It is also very helpful if you already know what the answer is using Computer Algebra Systems and thus you can plot a path towards that goal. You specified

without partial fractions or integration by parts

but these kind of tools are always in the back of the mind. They help to guide the thought process but they do not need to be written down explicitly. You may be able to detect hints of partial fractions and integration by parts even if they are not made explicit.

  • $\begingroup$ How exactly did you go through this thought process? Is it due to practice of similar questions or is there some key insight which I'm overlooking? $\endgroup$ – user671231 Jul 15 '19 at 18:26
  • $\begingroup$ Thanks a lot for the updated answer $\endgroup$ – user671231 Jul 18 '19 at 18:09
  • $\begingroup$ I'd also request you to show the solving for obtaining t I'm getting stuck in it $\endgroup$ – user671231 Jul 18 '19 at 18:22

$\newcommand{\d}{\mathrm{d}}$ I don't think there are any simple ways to evaluate this integral that aren't ad hoc.

Write the expression to the right of the integral sign as $$\omega=w^2(x^2 + x)\d x$$ where $w^{-1}=\mathrm{e}^x + x + 1$. Write $$\alpha=w^2x\d x - w \d x - \d w\text{.}$$ Note that the differential equation that $w$ satisfies is $\alpha = 0$. Our heuristic is to subtract off multiples of $\alpha$ from $\omega$ to reduce the degree until we get something that "looks integrable". Then we get $$\omega=\left(x+1+\tfrac{1}{w}\right)\alpha + \beta $$ where $$\begin{split}\beta &= \tfrac{\d w}{w}+ \d x + \d w + x \d w + w\d x\\ &= \d\left(\ln w + x + w + wx\right)\end{split}$$ Substituting in $w$, we get $$\int_0^x \frac{x^2 + x}{(\mathrm{e}^x+x+1)^2}\d x =-\ln(\mathrm{e}^x+x+1)+\ln 2 -\tfrac{1}{2}+x+\frac{1+x}{\mathrm{e}^x+ x + 1}$$

  • $\begingroup$ Thanks for sharing this answer. I haven't seen this way before though probably cause I'm beginning Integral calculus but it definitely seems to be quite simpler as compared to other methods like integration by parts $\endgroup$ – user671231 Jul 12 '19 at 20:41
  • $\begingroup$ How did you obtain the expression for $\omega$ in terms of $\alpha$ and $\beta$? $\endgroup$ – Zach Favakeh Jul 12 '19 at 21:12

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