$$ I _ { 1 } = \int \frac { x ^ { 4 } d x } { \sqrt { 2 - 3 x ^ { 10 } } } , $$

Question

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To evaluate the integral:

$$
I_1 = \int \frac{x^4 \, dx}{\sqrt{2 - 3x^{10}}}
$$

we can use a substitution method to simplify it.

**Step 1: Choose a Suitable Substitution**

Let’s set:

$$
u = x^5
$$

Then, calculate the differential $$ du $$:

$$
du = 5x^4 \, dx \quad \Rightarrow \quad x^4 \, dx = \frac{du}{5}
$$

**Step 2: Rewrite the Integral in Terms of $$ u $$**

Substituting $$ u = x^5 $$ into the integral:

$$
I_1 = \int \frac{x^4 \, dx}{\sqrt{2 - 3x^{10}}} = \int \frac{\frac{du}{5}}{\sqrt{2 - 3u^2}} = \frac{1}{5} \int \frac{du}{\sqrt{2 - 3u^2}}
$$

**Step 3: Evaluate the Integral**

The integral $$ \int \frac{du}{\sqrt{2 - 3u^2}} $$ is a standard form and can be recognized as:

$$
\int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin\left(\frac{u}{a}\right) + C
$$

In our case:

$$
a^2 = \frac{2}{3} \quad \Rightarrow \quad a = \sqrt{\frac{2}{3}}
$$

Thus,

$$
\int \frac{du}{\sqrt{2 - 3u^2}} = \frac{1}{\sqrt{3}} \arcsin\left(\frac{\sqrt{3}u}{\sqrt{2}}\right) + C
$$

**Step 4: Substitute Back to $$ x $$**

Replace $$ u $$ with $$ x^5 $$:

$$
I_1 = \frac{1}{5} \cdot \frac{1}{\sqrt{3}} \arcsin\left(\frac{\sqrt{3}x^5}{\sqrt{2}}\right) + C = \frac{1}{5\sqrt{3}} \arcsin\left(\frac{\sqrt{3}x^5}{\sqrt{2}}\right) + C
$$

**Final Answer:**

$$
I_1 = \frac{1}{5\sqrt{3}} \arcsin\left(\frac{\,\sqrt{3}\, x^{5}}{\sqrt{2}}\right) + C
$$

where $$ C $$ is the constant of integration.
The integral evaluates to: $$ I_1 = \frac{1}{5\sqrt{3}} \arcsin\left(\frac{\sqrt{3}x^5}{\sqrt{2}}\right) + C $$ where $$ C $$ is the constant of integration.

\( I _ { 1 } = \int \frac { x ^ { 4 } d x } { \sqrt { 2 - 3 x ^ { 10 } } } , \)

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