$I_{n}=\int_{-n}^{n}\{x+1\}\{x^2+2\}+\{x^2+2\}\{x^3+4\}dx$. Find $I_1$ where $\{\}$ denotes fractional part. $$I_{n}=\int_{-n}^{n}\{x+1\}\{x^2+2\}+\{x^2+2\}\{x^3+4\}dx$$. Find $I_1$ where $\{\}$ denotes fractional part.
$$I_1=\int_{-1}^{1}\{x+1\}\{x^2+2\}+\{x^2+2\}\{x^3+4\}dx$$
As $\{x+1\}=\{x\}$
$$I_1=\int_{-1}^{1}\{x\}\{x^2+2\}dx+\int_{-1}^{1}\{x^2+2\}\{x^3+4\}dx$$
$$I_1=I_2+I_3$$
$$I_2=\int_{-1}^{1}\{x\}\{x^2+2\}dx$$
$$I_2=\int_{-1}^{1}\{-x\}\{x^2+2\}dx$$
$$2I_2=\int_{-1}^{1}\{x^2+2\}\left(\{x\}+\{-x\}\right)dx$$
As $\{x\}+\{-x\}=1$ for $x\notin Z$. There are only finite number of integers in the interval $[-1,1]$, so we can ignore the area under them
$$2I_2=\int_{-1}^{1}\{x^2+2\}dx$$
$$I_2=\int_{0}^{1}\{x^2+2\}dx$$
$$I_2=\int_{0}^{1}x^2+2-\left[x^2+2\right]dx$$
$$I_2=\dfrac{1}{3}+2-2$$
$$I_2=\dfrac{1}{3}$$
$$I_3=\int_{-1}^{1}\{x^2+2\}\{x^3+4\}dx$$
$$I_3=\int_{-1}^{0}\{x^2+2\}\{x^3+4\}dx+\int_{0}^{1}\{x^2+2\}\{x^3+4\}dx$$
$$I_3=\int_{-1}^{0}\left(x^2+2-\left[x^2+2\right]\right)\left(x^3+4-\left[x^3+4\right]\right)dx+\int_{0}^{1}\left(x^2+2-\left[x^2+2\right]\right)\left(x^3+4-\left[x^3+4\right]\right)dx$$
$$I_3=\int_{-1}^{0}x^2\cdot (x^3+1)dx+\int_{0}^{1}x^5dx$$
$$I_3=\int_{-1}^{0}x^5dx+\int_{-1}^{0}x^2dx+\int_{0}^{1}x^5dx$$
$$I_3=\int_{-1}^{1}x^5dx+\int_{-1}^{0}x^2dx$$
$$I_3=0+\dfrac{1}{3}$$
$$I_1=I_2+I_3$$
$$I_1=\dfrac{2}{3}$$
But actual answer is $\dfrac{4}{3}$. What am I missing here?

Your calculations are correct. By employing a slightly different technique,
$$\int_0^1 \{x+1\}\{x^2+2\}dx= \int_0^1 \{x\}\{x^2\}dx= \int_0^1x^3dx = \frac 14;$$
$$\int_{-1}^0\{x+1\}\{x^2+2\}dx=\int_{-1}^0\{x+1\}\{(x+1)^2+1-2x\}dx=$$
$$\int_0^1\{x\}\{x^2+3-2x\}dx=\int_0^1\{x\}\{x^2+1-2x\}dx=\int_0^1\{x\}\{(1-x)^2\}dx=$$
$$\int_0^1x(1-x)^2dx = \frac{1}{12}.$$
Similarly,
$$\int_{-1}^0\{x^2+2\}\{x^3+4\}dx =\frac 16$$
and
$$\int^{1}_0\{x^2+2\}\{x^3+4\}dx =\frac 16.$$
In total this gives, as you already found, $\frac 23$.
