How do you integrate #int x(5^(-x^2))dx#?

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Answer 1 · 2016-11-28T07:59:06

The answer is #=-(1/2)5^(-x^2)/ln5+C#

We use the substitution

#u=-x^2#

#du=-2xdx#

#xdx=(-du)/2#

Therefore,

#intx(5^(-x^2))dx=-1/2int 5^(u)du#

Let, #y=5^(u)#

Then taking the logarithm

#lny=u ln5#

#y=e^(u ln5)#

#int 5^(u)du=inte^(u ln5)du=e^(u ln5)/ln5=y/ln5=5^(u)/ln5#

Therefore,

#intx(5^(-x^2))dx=-(1/2)5^(-x^2)/ln5+C#