How do you integrate $\int x (5^{- x^{2}}) d x$ $int x(5^(-x^2))dx$ ?

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How do you integrate $\int x (5^{- x^{2}}) d x$ $int x(5^(-x^2))dx$ ?

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Answer 1 · 2016-11-28T07:59:06

The answer is $= - (\frac{1}{2}) \frac{5^{- x^{2}}}{ln 5} + C$ $=-(1/2)5^(-x^2)/ln5+C$

Explanation:

We use the substitution

$u = - x^{2}$ $u=-x^2$

$d u = - 2 x d x$ $du=-2xdx$

$x d x = \frac{- d u}{2}$ $xdx=(-du)/2$

Therefore,

$\int x (5^{- x^{2}}) d x = - \frac{1}{2} \int 5^{u} d u$ $intx(5^(-x^2))dx=-1/2int 5^(u)du$

Let, $y = 5^{u}$ $y=5^(u)$

Then taking the logarithm

$ln y = u ln 5$ $lny=u ln5$

$y = e^{u ln 5}$ $y=e^(u ln5)$

$\int 5^{u} d u = \int e^{u ln 5} d u = \frac{e^{u ln 5}}{ln 5} = \frac{y}{ln 5} = \frac{5^{u}}{ln 5}$ $int 5^(u)du=inte^(u ln5)du=e^(u ln5)/ln5=y/ln5=5^(u)/ln5$

Therefore,

$\int x (5^{- x^{2}}) d x = - (\frac{1}{2}) \frac{5^{- x^{2}}}{ln 5} + C$ $intx(5^(-x^2))dx=-(1/2)5^(-x^2)/ln5+C$

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How do you integrate $\int x (5^{- x^{2}}) d x$ $int x(5^(-x^2))dx$ ?

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