The value of $\int\frac{x^5}{\sqrt{1+x^3}}dx$ is

$\frac{2}{9} (1+x^3)^{3/2} -\frac{2}{3} (1+x^3)^{\frac{1}{2}} + c$: $c$ is an arbitrary constant

The correct answer is Option (1) → $\frac{2}{9} (1+x^3)^{3/2} -\frac{2}{3} (1+x^3)^{\frac{1}{2}} + c$: $c$ is an arbitrary constant

Given: $\int \frac{x^5}{\sqrt{1 + x^3}} \, dx$

Use substitution: $x^3 = t \Rightarrow 3x^2 dx = dt \Rightarrow x^2 dx = \frac{dt}{3}$

$x^5 = x^3 \cdot x^2 = t \cdot x^2$

So the integral becomes:

$\int \frac{t \cdot x^2}{\sqrt{1 + t}} \cdot dx = \int \frac{t}{\sqrt{1 + t}} \cdot \frac{dx}{x^2}$

But $x^2 dx = \frac{dt}{3}$, so $dx = \frac{dt}{3x^2}$

Instead, express all in terms of $t$:

From $x^3 = t$, so $x = t^{1/3}$ and $x^2 = t^{2/3}$

Then $x^5 = t \cdot t^{2/3} = t^{5/3}$, and $dx = \frac{1}{3}t^{-2/3} dt$

So, $\int \frac{x^5}{\sqrt{1 + x^3}} dx = \int \frac{t^{5/3}}{\sqrt{1 + t}} \cdot \frac{1}{3}t^{-2/3} dt$

$= \frac{1}{3} \int \frac{t^{1}}{\sqrt{1 + t}} dt$

$= \frac{1}{3} \int \frac{t}{\sqrt{1 + t}} dt$

Now use substitution: $u = 1 + t \Rightarrow du = dt$

$t = u - 1$

So the integral becomes:

$\frac{1}{3} \int \frac{u - 1}{\sqrt{u}} du = \frac{1}{3} \int (u^{1/2} - u^{-1/2}) du$

$= \frac{1}{3} \left( \frac{2}{3} u^{3/2} - 2 u^{1/2} \right) + C$

$= \frac{1}{3} \left( \frac{2}{3}(1 + x^3)^{3/2} - 2(1 + x^3)^{1/2} \right) + C$