Answer
$$
\int (x^3+1)\cos (x^4+4x ) dx
=\frac{1}{4} \sin (x^4+4x ) +c.
$$
Work Step by Step
Since $ u=x^4+4x $, then $ du=4(x^3+1)dx $ and hence, $$
\int (x^3+1)\cos (x^4+4x ) dx=\frac{1}{4}\int \cos u d u=\frac{1}{4} \sin u+c\\
=\frac{1}{4} \sin (x^4+4x ) +c.
$$
