Result misses a "-" sign

Question

$$ \int \frac{x^{n+2}}{e^{x^2}(n+1)} = \frac{1}{n+1} \int x^{n+2}e^{-x^2}dx \to n+2=u, n+1=u-1 $$ $$ \frac{1}{u-1}\int x^ue^{-x^2}dx \to x^2=k, x=\sqrt{k}, dt=\frac{1}{2\sqrt{k}}dk $$ $$ \frac{1}{2(u-1)}\int k^{u/2}e^{-k}k^{-1/2}dk = \frac{1}{2(u-1)}\int k^{\frac{u-1}{2}}e^{-k}dk = \frac{1}{2(n+1)}\int k^{\frac{n+1}{2}}e^{-k}dk = I $$ $$ \to -\Gamma\biggl(\frac{n+3}{2},k\biggl) = \int_{k}^{\infty} t^{\frac{n+3}{2}-1}e^{-t}dt = \int_{k}^{\infty} t^{\frac{n+1}{2}}e^{-t}dt \to I = -\frac{1}{2(n+1)}\Gamma\biggl(\frac{n+3}{2},x^2\biggl)$$ $$ \to a\int x^ne^{-x^2}dx = a\biggl[\frac{x^{n+1}}{e^{x^2}(n+1)} + \frac{1}{(n+1)}\Gamma\biggl(\frac{n+3}{2},x^2\biggl)\biggl] $$ Apparently the result should be $$ \to a\int x^ne^{-x^2}dx = a\biggl[\frac{x^{n+1}}{e^{x^2}(n+1)} - \frac{1}{(n+1)}\Gamma\biggl(\frac{n+3}{2},x^2\biggl)\biggl] $$ but i can't find any mistakes, if anyone could help it'd be great!

Answer 1

Note that $$ \frac{\partial}{\partial x} \Gamma(s, x) = \color{red}{-} x^{s - 1} e^{-x}, $$ not $$ \color{red}{+} x^{s - 1} e^{-x} $$ since the variable $x$ is in the lower limit, not the upper limit, of the integral.

Hence the expression for $I$ in the fourth line in terms of the incomplete Gamma function is off by a minus sign.