Standard normal distribution find $\mathbb{E}X(X+1)$ and $\mathbb{E}e^{3X^2/8}$

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Random variable X has standard normal distribution. Find $\mathbb{E}X(X+1)$ and $\mathbb{E}e^{3X^2/8}$. My attempt is to solve: $\int_{-\infty}^{\infty}x(x-1) \frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}$ and $\int_{-\infty}^{\infty}e^{3x^2/8} \frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}=2\int_{-\infty}^\infty \frac{1}{2\sqrt{2\pi}}e^{\frac{-x^2}{8}}=2$.

Thus my question is what to do with the first example (how to calculate this integral?) and is my answer correct for the second?

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$$\int x^2e^{\frac{-x^2}{2}}=\int x\cdot xe^{\frac{-x^2}{2}}=-x\cdot e^{\frac{-x^2}{2}}+\int e^{\frac{-x^2}{2}}$$ $$\int xe^{\frac{-x^2}{2}}=-e^{\frac{-x^2}{2}}$$ $$\therefore \int (x^2-x)e^{\frac{-x^2}{2}}=-(x-1)\cdot e^{\frac{-x^2}{2}}+\int e^{\frac{-x^2}{2}}$$

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$$\mathbb{E}[X^2+X]=\mathbb{E}[X^2]+\mathbb{E}[X]=\mathbb{E}[(X-\mathbb{E}[X])^2]+\mathbb{E}[X]=\sigma^2 +0=1$$