#int_{-infty}^0 x5^{-x^2} dx#

by the substitution #u=-x^2#.

#Rightarrow {du}/{dx}=-2x Rightarrow dx={du}/{-2x}#

As #x# goes from #-infty# to #0#, #u# goes from #-infty# to #0#.

#=int_{-infty}^0x5^{u}{du}/{-2x}#

by cancelling out #x#'s and pulling #-1/2# out of the integral,

#=-1/2int_{-infty}^0 5^u du#

by the definition of improper integral,

#=-1/2lim_{t to -infty}int_t^0 5^u du#

by finding an anti-derivative,

#=-1/2lim_{t to -infty}[{5^u}/{ln5}]_t^0#

by pulling #1/{ln5}# out,

#=-1/{2ln5}lim_{t to -infty}(1-5^t)#

since #lim_{t to -infty}5^t=0#,

#=-1/{2ln5}#

Hence, the improper integral converges to #-1/{2ln5}#.

I hope that this was helpful.