How to integrate $\int e^{(x+1)^{2}}(6x+23)dx$?

Question

How to integrate $$\int e^{(x+1)^{2}}(6x+23)dx$$ ? I am trying this question by multiplying the numerator and denominator by $(2x+2)$. Therefore, the above integral will become $$\int \frac{e^{(x+1)^{2}}(2x+2)(6x+23)}{(2x+2)}dx$$. Now Let's substitute $(x+1)^{2}=u$. Therefore, $(2x+2)dx=du$. Therefore the above integral will become $$\int \frac{e^{u}(6(\pm{\sqrt{u}}-1)+23)}{(2(\pm{\sqrt{u}}-1)+2)}du$$. Therefore, finally after simplifying a bit, the above integral will become $$\int \frac{e^{u}(\pm{6\sqrt{u}}+17)}{(\pm{2\sqrt{u}})}du$$. But after this step, I can't approach further. Please help me out with this integral.

Answer 1

$\int e^{({x+1})^2}(6x+23)dx$

$\int e^{({x+1})^2}6x \thinspace dx+ 23\int e^{({x+1})^2}\thinspace dx$

$6\int e^{({x+1})^2}(x+1) \thinspace dx-6\int e^{({x+1})^2}\thinspace dx+ 23\int e^{({x+1})^2}\thinspace dx$

$3\int e^{u} \thinspace du+17\int e^{({x+1})^2}\thinspace dx$

$3e^{{(x+1)}^2} +\frac{17 \mathrm{erfi}(x+1)\sqrt\pi}{2}+C$