<s>[INST] $f\left( x \right) = {x^3} + x$, then $\int\limits_1^2 {f\left( x \right)dx} + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $ 
If $f\left( x \right) = {x^3} + x$, then
$$\int\limits_1^2 {f\left( x \right)dx}  + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $$
is________.

My approach is as follows:
$$g = {f^{ - 1}} \Rightarrow g\left( x \right) = {f^{ - 1}}\left( x \right)$$
$$g\left( {2x} \right) = {f^{ - 1}}\left( {2x} \right)$$
$$2y = {8x^3} + 2x$$
$${f^{ - 1}}\left( {{x^3} + x} \right) = x$$
$$\int\limits_1^2 {f\left( x \right)dx}  + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $$
I am not able to proceed further.
 [/INST] After the substitution $u=2x$ in the second integral, your expresion becomes
$$\int\limits_1^2 {f\left( x \right)dx}  + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx}= \int\limits_1^2 {f\left( x \right)dx} + \int\limits_{f(1)=2}^{f(2)=10} {{f^{ - 1}}\left( {u} \right)du}$$
Now, you can use an integration rule for the inverse function and obtain
$$\int\limits_1^2 {f\left( x \right)dx} + \int\limits_{f(1)=2}^{f(2)=10} {{f^{ - 1}}\left( {u} \right)du} = 2\cdot f(2) - 1\cdot f(1) =18$$
</s>