MHB Integrating exponential functions

[karush]

$\int x (5^{-x^2})dx= -\frac{1}{2} \int 5^{-x^{2}}(-2x)dx$

how is $\frac{1}{2}$ in front of the $\int$ derived

 

[JJacquelin]

dx is missing !
5^-x²=exp(-x²ln(5))
x²=t
x dx = dt/2

 

[karush]

sorry i hit the save instead of preview thot I cud fix it before reply

 

[MarkFL]

I've done that before, and it is annoying! (Headbang)

 

[CaptainBlack]

$\int x (5^{-x^2})dx= -\frac{1}{2} \int 5^{-x^{2}}(-2x)dx$

how is $\frac{1}{2}$ in front of the $\int$ derived

\[\int x (5^{-x^2})dx= \int x e^{-x^2\ln(5)}dx\]

Now: \[\frac{d}{dx} e^{-x^2\ln(5)}=-2x\ln(5) e^{-x^2\ln(5)}\] so:

\[\begin{aligned}\int x (5^{-x^2})dx &=-\frac{1}{2\ln(5)}\int \frac{d}{dx} e^{-x^2\ln(5)} dx\\
&=-\frac{e^{-x^2\ln(5)}}{2\ln(5)}+ C \\ &=-\;\frac{5^{-x^2}}{2\ln(5)}+C \end{aligned} \]

CB