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Which is the right way to solve indefinite integrals which contain absolute values? For example if I have $\int |2x+3| e^x dx$
Can I consider the sign function and integrate separetly? I mean doing: $ Sign(2x+3) \int (2x+3)e^x dx$

Or maybe I should use the definition of absolute value and divide the two possibilities
$\int (2x+3)e^x dx$ if $ (2x+3)>0$ and $\int (-2x-3)e^x dx$ if $ (2x+3)<0$

But I think that's more suitable for definite rather than indefinite integrals

How can I solve this type of integrals? Thanks a lot in advice

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    $\begingroup$ Since the antiderivative is linear, your answer would differ by at most a sign change. This would matter when you are plugging an antiderivate into, say, limits to calculate a definite integral. $\endgroup$
    – amcalde
    Commented Nov 17, 2015 at 19:40

2 Answers 2

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HINT: i would consider the cases $$2x+3\geq 0$$ or $$2x+3<0$$

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$\int|2x+3|e^x~dx$

$=\text{sgn}(2x+3)\int(2x+3)e^x~dx$

$=\text{sgn}(2x+3)\int_{-\frac{3}{2}}^x(2x+3)e^x~dx+C$

$=\text{sgn}(2x+3)\int_{-\frac{3}{2}}^x(2x+3)~d(e^x)+C$

$=\text{sgn}(2x+3)[(2x+3)e^x]_{-\frac{3}{2}}^x-\text{sgn}(2x+3)\int_{-\frac{3}{2}}^xe^x~d(2x+3)+C$

$=\text{sgn}(2x+3)(2x+3)e^x-2~\text{sgn}(2x+3)\int_{-\frac{3}{2}}^xe^x~dx+C$

$=|2x+3|e^x-2~\text{sgn}(2x+3)[e^x]_{-\frac{3}{2}}^x+C$

$=|2x+3|e^x-2~\text{sgn}(2x+3)\left(e^x-e^{-\frac{3}{2}}\right)+C$

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  • $\begingroup$ Thanks a lot for this answer, I just did not understand why you used the integral function, can't I just integrate indefinetly? $\endgroup$
    – Gianolepo
    Commented Nov 20, 2015 at 16:56
  • $\begingroup$ @Francesco Caruso $\text{sgn}(2x+3)e^x$ is discontinuous at $x=-\dfrac{3}{2}$ , while $\text{sgn}(2x+3)\left(e^x-e^{-\frac{3}{2}}\right)$ is continuous at $x=-\dfrac{3}{2}$ . $\endgroup$
    – Harry Peter
    Commented Dec 4, 2015 at 16:00

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