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Solution
Your input: calculate $$$\int_{-1}^{1}\left( \ln{\left(x^{3} \right)} \right)dx=\int_{-1}^{1}\left( 3 \ln{\left(x \right)} \right)dx$$$
First, calculate the corresponding indefinite integral: $$$\int{3 \ln{\left(x \right)} d x}=3 x \left(\ln{\left(x \right)} - 1\right)$$$ (for steps, see indefinite integral calculator)
According to the Fundamental Theorem of Calculus, $$$\int_a^b F(x) dx=f(b)-f(a)$$$, so just evaluate the integral at the endpoints, and that's the answer.
$$$\left(3 x \left(\ln{\left(x \right)} - 1\right)\right)|_{\left(x=1\right)}=-3$$$
$$$\left(3 x \left(\ln{\left(x \right)} - 1\right)\right)|_{\left(x=-1\right)}=3 - 3 i \pi$$$
$$$\int_{-1}^{1}\left( 3 \ln{\left(x \right)} \right)dx=\left(3 x \left(\ln{\left(x \right)} - 1\right)\right)|_{\left(x=1\right)}-\left(3 x \left(\ln{\left(x \right)} - 1\right)\right)|_{\left(x=-1\right)}=-6 + 3 i \pi$$$
Answer: $$$\int_{-1}^{1}\left( \ln{\left(x^{3} \right)} \right)dx=-6 + 3 i \pi\approx -6.0 + 9.42477796076938 i$$$