∫xexdx
One of my friends said substitution , but I can't seem to get it to work.
Otherwise I also tried integration by parts but I'm not getting the same answer as wolfram.
The space in the question seems like it shouldn't take more than 2 lines though. Am I missing something?
Thanks to all the answers below , I messed up in the original question it was actually
∫xex2dx
With help from the below answers I did the following:
Let u=x2 , then du=2xdx
So rewriting the integral
∫x⋅eu12xdx
Simplifying yields:
12x∫eudx
Which in turn yields:
eu2+C
The rest is fairly obvious!
Answer
Definitely by parts, as substitution of x won't get you anywhere.
Let u=ex and dv=ex in ∫udv=uv−∫vdu
and we have
∫xexdx=xex−∫exdx=xex−ex+c
As a general tip, you will usually want to use parts if you have an exponential, which doesn't get any nastier as you anti-differentiate, and a polynomial which will disappear after some number of differentiations. A notable exception is when you have something like
∫xex2dx
Where a u=x2 substitution will cancel out the x coefficient on the exponential.
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