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,

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

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