Fnd $ \int\frac{e^{-2x-x^2}}{( x+1)^2}\,dx$

Question

Find $\displaystyle\int \dfrac{e^{-2x-x^2}}{\left( x+1\right)^2}\hspace{1mm}dx$.

If I do Integration by parts, I end up with $\displaystyle\int e^{-2x-x^2}\hspace{1mm}dx$

Which I believe cannot be integrated

Answer 1

Since $$\int \frac{e^{-2x-x^2}}{(x+1)^2}\,dx = e\cdot\int \frac{e^{-(x+1)^2}}{(x+1)^2}$$ your question is equivalent to finding a primitive for $\frac{e^{-x^2}}{x^2}$. Integration by parts gives: $$\int \frac{e^{-x^2}}{x^2} = -\frac{e^{-x^2}}{x}-\sqrt{\pi}\cdot\operatorname{Erf}(x).$$

Answer 2

Notice that $$\int \frac{e^{(-2x-x^2)}}{(x+1)^2}\,dx=\int e\cdot\frac{e^{(-1-2x-x^2)}}{(x+1)^2}\,dx=\int e\frac{e^{-(x+1)^2}}{(x+1)^2}\,d(x+1)$$ Substitute $y=x+1$ to get (using integration by parts) $$\int e\frac{e^{-y^2}}{y^2}\,dy=e\int \frac{e^{-y^2}}{y^2}\,dy=e\Big(-\frac{e^{-y^2}}{y}+\int\frac{-2ye^{-y^2}}{y}\,dy\Big)=e\Big(-\frac{e^{-y^2}}{y}-2\int e^{-y^2}\,dy\Big)=e\Big(-\frac{e^{-y^2}}{y}-\sqrt{\pi}\cdot erf(y)\Big)$$ Substitute back $y=x+1$ to get the desired result.