Showing that $\int_x^{\infty}\frac{1}{u^2}e^{(-u^2/2)}du=\frac{1}{x}e^{(-x^2/2)}-\int_x^{\infty}e^{{(-u^2/2)}}du$

Question

My texbook claims that integration by parts of the integral $\int_x^{\infty}\frac{1}{u^2}e^{\frac{-u^2}{2}}du$, with the hint that $d(-\frac{1}{u})=\frac{du}{u^2}$, gives $$\frac{1}{x}e^{\frac{-x^2}{2}}-\int_x^{\infty}e^{\frac{-u^2}{2}}du$$

Now when I try to express $\int sdv$ as $sv-\int vds$. I get $s=-\frac{1}{u}$, $ds=\frac{du}{u^2}$ then $$dv=\int-\frac{1}{u}e^{\frac{-u^2}{2}}du$$ which i cannot integrate. Whats wrong here?

Thanks in advance!

Answer 1

Hint: Try $s = e^\frac{-u^2}{2}$, and $dv = \frac{1}{u^2}$

Answer 2

Just do the opposite : s = Exp[-u^2/2] , dv = du / u^2. Then ds =- u Exp[-u^2/2] , v = - 1/ u. Then built the antiderivative and use the bounds (and limits).