Answer
$$
\int \frac{t^3}{(4-2t^4)^{11}} dx =\frac{1}{80}\frac{1}{(4-2t^4)^{10}}+c.
$$
Work Step by Step
Since $ u=4-2t^4$, then $ du=-8t^3dt $ and hence, $$
\int \frac{t^3}{(4-2t^4)^{11}} dx=-\frac{1}{8}\int \frac{1}{u^{11}} d u=-\frac{1}{8}\int u^{-11} d u\\
=-\frac{1}{8} \frac{-1}{10}u^{-10} +c=\frac{1}{80}\frac{1}{(4-2t^4)^{10}}+c.
$$
