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 #