What is $int (8x)/e^(x^2) \ dx$ ?

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Answer 1 · 2017-09-26T14:58:33

I got: $-4e^(-x^2)+c$

Have a look:
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Answer 2 · 2017-09-26T15:00:04

$int (8x)/e^(x^2) \ dx = -4 e^(-x^2) + c$

We seek:

$I = int (8x)/e^(x^2) \ dx$

Which we can write as:

$I = -4 \ int (-2x)e^(-x^2) \ dx$

With practice we can integrate this directly, but for the benefit of those unable to do this, let us perform substitution:

Let $u=e^(-x^2) => (du)/dx = -2xe^(-x^2)$

So if we now substitute this into the integral, we get:

$I = -4 \ int \ du$

Which is now a standard integral, so we have:

$I = -4 u + c$

And reversing the substitution:

$I = -4 e^(-x^2) + c$

What is int (8x)/e^(x^2) \ dxint (8x)/e^(x^2) \ dx?