Solve integral without partial fractions or integration by parts 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.
 A: 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.
A: $\newcommand{\d}{\mathrm{d}}$
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
$$\int_0^x \frac{x^2 + x}{(\mathrm{e}^x+x+1)^2}\d x =-\ln(\mathrm{e}^x+x+1)+\ln 2 -\tfrac{1}{2}+x+\frac{1+x}{\mathrm{e}^x+ x + 1}$$
