To solve the integral $\int x^{5} \sqrt{x^{3}+1} \, dx$, we can use a substitution method. Let's set $u = x^3 + 1$, which simplifies the integral. Then, we find $du$ in terms of $dx$ and substitute back into the integral to solve it.
We are given the integral:
\[
\int x^{5} \sqrt{x^{3}+1} \, dx
\]
To simplify this integral, we can use substitution. Let:
\[
u = x^3 + 1
\]
Then, the derivative of \(u\) with respect to \(x\) is:
\[
\frac{du}{dx} = 3x^2 \implies du = 3x^2 \, dx \implies \frac{du}{3} = x^2 \, dx
\]
Next, we need to express \(x^5 \sqrt{x^3 + 1} \, dx\) in terms of \(u\). Notice that:
\[
x^5 = x^3 \cdot x^2 = (u - 1) \cdot x^2
\]
Thus, the integral becomes:
\[
\int x^5 \sqrt{x^3 + 1} \, dx = \int (u - 1) x^2 \sqrt{u} \, dx
\]
Substitute \(x^2 \, dx\) with \(\frac{du}{3}\):
\[
\int (u - 1) \sqrt{u} \cdot \frac{du}{3}
\]
Factor out the constant \(\frac{1}{3}\):
\[
\frac{1}{3} \int (u - 1) \sqrt{u} \, du
\]
Distribute \(\sqrt{u}\) inside the integral:
\[
\frac{1}{3} \int (u^{3/2} - u^{1/2}) \, du
\]
Now, integrate each term separately:
\[
\frac{1}{3} \left( \int u^{3/2} \, du - \int u^{1/2} \, du \right)
\]
Using the power rule for integration:
\[
\int u^{3/2} \, du = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2}
\]
\[
\int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}
\]
Thus, the integral becomes:
\[
\frac{1}{3} \left( \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} \right)
\]
Combine the terms:
\[
\frac{1}{3} \left( \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} \right) = \frac{2}{15} u^{5/2} - \frac{2}{9} u^{3/2}
\]
Finally, substitute back \(u = x^3 + 1\):
\[
\frac{2}{15} (x^3 + 1)^{5/2} - \frac{2}{9} (x^3 + 1)^{3/2} + C
\]
\[
\boxed{\frac{2}{15} (x^3 + 1)^{5/2} - \frac{2}{9} (x^3 + 1)^{3/2} + C}
\]
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