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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.

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    $\begingroup$ $\int (2x+9)e^x dx = (2x+7)e^x + C$. So, even though your question has a lot of typos, your final answer appears to match this result. $\endgroup$
    – Lord Soth
    Jun 27 '13 at 17:19
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    $\begingroup$ No, it's not wrong...unless you mean the integration constant's missing. $\endgroup$
    – DonAntonio
    Jun 27 '13 at 17:19
  • $\begingroup$ But I have +9 and you guys have +7 $\endgroup$
    – user84004
    Jun 27 '13 at 17:28
  • $\begingroup$ @user84004: factor out the $e^x$ to get $e^x (2x + 9 - 2) = e^x ( 2x + 7)$. $\endgroup$ Jun 27 '13 at 17:38
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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.

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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.

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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.

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  • $\begingroup$ not sure why this didn't get a TU - has it now! +1 $\endgroup$
    – Amzoti
    Jun 28 '13 at 2:29

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