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I integrated $e^x\sinh(x)$ by writing out $\sinh x$ as $\frac{e^x-e^{-x}}{2}$ and multiplying to get

$\int$$\frac{e^{2x}-e^{0}}{2}dx$

which comes out to be

$\frac{e^{2x}}{4}- \frac{x}{2}$ + C

however, plugging the question into photo math they use integration by parts and the answer is instead:

$\frac{e^{2x}-2}{4}- \frac{x}{2}$ + C

I was wondering which answer is correct and why does solving it in different ways give different answers?

Sine of the Time
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user1139725
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    note that the answers differ by a constant – Sine of the Time Nov 04 '23 at 22:55
  • $\frac{e^{2x}-2}{4}-\frac x2+C=\frac{e^{2x}}{4}-\frac12-\frac x2+C$ – Sine of the Time Nov 04 '23 at 22:56
  • It's definitely possible to solve an indefinite integral in various ways and get answers that "look" different, but the constant of integration would make those "different answers" agree. You could also differentiate your answer and see if you get the original integrand back. – Accelerator Nov 05 '23 at 01:00

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