Answer
$$\frac{2}{9} (1+x^3)^{3/2}+C$$
Work Step by Step
Given $$\int \sqrt{x^{4}+x^{7}} d x$$
Since
\begin{align*}
\int \sqrt{x^{4}+x^{7}} d x&=\int \sqrt{x^{4}(1+x^{3})} d x\\
&=\int x^2\sqrt{1+x^3}dx
\end{align*}
Let $$u=1+x^3,\ \ \ \ du=3x^2dx$$
Then
\begin{align*}
\int \sqrt{x^{4}+x^{7}} d x&=\int \sqrt{x^{4}(1+x^{3})} d x\\
&=\int x^2\sqrt{1+x^3}dx\\
&=\frac{1}{3} \int u^{1/2}du\\
&=\frac{2}{9} u^{3/2}+C\\
&=\frac{2}{9} (1+x^3)^{3/2}+C
\end{align*}
