$f\left( x \right) = {x^3} + x$, 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.

$$f(x)=x^3+x$$
Let $$K=\int_{1}^{2} f(x) dx+2\int_{1}^{5} f^{-1}(2x) dx=I+J$$
$$ I=\int_{1}^{2} (x^3+x) dx=\frac{21}{4}$$
$$J=2\int_{1}^{5} f^{-1}(2x)dx=\int_{2}^{10} f^{-1}(z) dz$$
Let $f^{-1}(z)=t \implies z=f(t) \implies dz=f'(t) dt$, then
$$J=\int_{1}^{2} t f'(t) dt=\int_{1}^{2}t(3t^2+1)dt=\frac{51}{4}.$$
Finally $$K=21/4+51/4=18.$$
