Indefinite integral of $(2x+9)e^x$ 
What is the indefinite integral $\displaystyle\int (2x+9)e^x\,\mathrm dx$?

Attempt:
Integration by parts seems obvious.
$u = 2x + 9, \mathrm du = 2$
$\mathrm dv = e^x, v = e^x$
$uv - \int v\,\mathrm du$
$(2x+9)e^x - \int 2e^x$
$(2x+9)e^x - 2e^x$
This is wrong but I don't see why.
 A: Check:
$$
\frac{d[(2x+7)e^x]}{dx}
  = e^x \frac{d[2x+7]}{dx} + (2x+7) \frac{d[e^x]}{dx}
  = 2e^x + (2x+7)e^x
  = (2x+9)e^x
$$
as desired, so your answer is correct. Just add the integration constant if you like.
A: Your method is just fine, and your result (if you add $+C$) is correct. Perhaps you are comparing your solution to a solution manual:
Note that the manual may have factored out $e^x$ in your result: 
$$ (2x+9)e^x - 2e^x +  + C = (2x + 9 - 2)e^x + C = (2x+7)e^x + C$$
...but your expression (left-hand side of the equation above) is equivalent to the right-hand side of the equation.
A: As the comment mention, the result is not wrong; we can verify this by differentiation (using the product rule):
$$\frac{\mathrm d}{\mathrm dx}(2x+7)e^x = (2x+7)e^x + 2e^x = (2x+9)e^x$$
(Note that $(2x+9)e^x - 2e^x = ((2x+9)-2))e^x = (2x+7)e^x$.)

However, it is important to stress that for a function $u = u(x)$, one has:
$$\mathrm d u = u'(x) \,\mathrm dx$$
as opposed to simply $u'(x)$. When you get further in your mathematical studies, you will get to know that:


*

*$\mathrm dx$ is a thing called a $1$-form;

*For functions $u(x_1,x_2,\ldots,x_n)$, we have $\mathrm du = \dfrac{\partial u}{\partial x_1} \,\mathrm dx_1 +\ldots+\dfrac{\partial u}{\partial x_n}\,\mathrm dx_n$;


I hope that instigates some curiosity as to what these mysterious $\mathrm dx$s are, and what the intuition behind them should be.
