Can you help me finish this problem? Decimal approximations are not sufficient for credit. I wasn't sure how to simplify the left side much more (besides obviously multiplying by two and subtracting the one over) to make it look something similar enough to make a direct comparison with the right side.
$$ \begin{aligned} &\int_{0}^{\sqrt{\sqrt{7}-1}}\left[\left(x^{3}+x\right) e^{-x^{2}}\right] d x \leq \ln (2) \\ &\int_{0}^{\sqrt{\sqrt{7}-1}}\left(e^{-x^{2}} x^{3}+e^{-x^{2}} x\right) d x \\ &\int_{0}^{\sqrt{\sqrt{7}-1}} e^{-x^{2}} x^{3} d x+\int_{0}^{\sqrt{\sqrt{7}-1}} e^{-x^{2}} x d x \end{aligned} $$ $$ \begin{aligned} & \text { Let } u=-x^{2} ; \text { Let } d u=-2 x \\ & -\int_{0}^{\sqrt{\sqrt{7}-1}} \frac{e^{u} u}{2} d u-\frac{1}{2} \int_{0}^{\sqrt{\sqrt{7}-1}} e^{u} d u \\ & \frac{x^{2} e^{-x^{2}}}{2}-\frac{e^{-x^{2}}}{2} \mid \sqrt{\sqrt{7}-1} \\ & \frac{e^{-x^{2}}\left(x^{2}-1\right)}{2} \mid \begin{gathered}\sqrt{\sqrt{7}-1} \\0\end{gathered} \end{aligned} $$ $$ \begin{gathered} \frac{e^{-\sqrt{\sqrt{7}-1}^{2}}\left(\sqrt{\sqrt{7}-1}^{2}-1\right)}{2}-\frac{e^{0}\left(0^{2}-1\right)}{2} \leq \ln (2) \\ \frac{e^{-(\sqrt{7}-1)}(\sqrt{7}-1-1)}{2}+\frac{1}{2} \leq \ln (2) \\ \frac{e^{-\sqrt{7}+1}(\sqrt{7}-2)+1}{2} \leq \ln (2) \\ \approx 0.62 \leq 0.69 \end{gathered} $$
Using almost what you did $$\int_0^a (x^3+x)\,e^{-x^2}\,dx=1-\frac{1}{2} \left(a^2+2\right) e^{-a^2}$$ which increases with $a$.
Solving for the equality, let $a^2=b$ $$1-\frac{1}{2} \left(b+2\right) e^{-b}=\log(2) \implies b=-2-W_{-1}\left(\frac{2 (\log (2)-1)}{e^2}\right)$$ where $W_{-1}(t)$ is the second branch of Lambert function.
Using the series expansion given in the liked Wikipedia page $$W_{-1}(z)= L_1-L_2+\frac {L_2}{L_1}+\frac {L_2(L_2-2)}{2L_1^2}+\cdots$$ with $L_1=\log(-z)$ and $L_2=\log(-L_1)$ gives $$W_{-1}\left(\frac{2 (\log (2)-1)}{e^2}\right)=-3.8315 \implies b=1.8315\implies a=1.3533$$ while $\sqrt{\sqrt{7}-1}=1.28287$.
Brute force would be to make $a=\sqrt{\sqrt{7}-1}$ in $$1-\frac{1}{2} \left(a^2+2\right) e^{-a^2}=1-\frac{1}{2} \left(1+\sqrt{7}\right) e^{1-\sqrt{7}}=0.64843 < \log(2)$$
Using the first part, you could even show that $$\int_0^{ \sqrt{\sqrt{8}-1}} (x^3+x)\,e^{-x^2}\,dx< \log(2)$$