`int(x dx)/((x^2 + 1)^(4//5)(x^2 + 2)^(6//5))` =

Options

A. `2/5 ((x^2 + 2)/(x^2 + 1))^(1//5)`
B. `5/2 ((x^2 + 2)/(x^2 + 1))^(-1//5)`
C. `5/2 ((x^2 + 1)/(x^2 + 2))^(1//5)`
D. none of these

Answer

To solve the integral

\[ I = \int \frac{x \, dx}{(x^2 + 1)^{\frac{4}{5}} (x^2 + 2)^{\frac{6}{5}}} \]

we can follow these steps:

### Step 1: Rewrite the Integral
We can separate the powers in the denominator:

\[ I = \int \frac{x \, dx}{(x^2 + 2)^{\frac{6}{5}} (x^2 + 1)^{\frac{4}{5}}} = \int \frac{x \, dx}{(x^2 + 2)^{\frac{4}{5}} (x^2 + 1)^{\frac{4}{5}} (x^2 + 2)^{\frac{2}{5}}} \]

### Step 2: Substitution
Let

\[ t = \frac{x^2 + 1}{x^2 + 2} \]

Then, we differentiate \(t\):

\[ dt = \frac{(2x)(x^2 + 2) - (2x)(x^2 + 1)}{(x^2 + 2)^2} \, dx = \frac{2x}{(x^2 + 2)^2} \, dx \]

Thus,

\[ dx = \frac{(x^2 + 2)^2}{2x} \, dt \]

### Step 3: Substitute in the Integral
Substituting \(t\) and \(dx\) into the integral:

\[ I = \int \frac{x \cdot \frac{(x^2 + 2)^2}{2x} \, dt}{(x^2 + 2)^{\frac{6}{5}} (x^2 + 1)^{\frac{4}{5}}} \]

This simplifies to:

\[ I = \frac{1}{2} \int \frac{(x^2 + 2)^{2 - \frac{6}{5}}}{(x^2 + 1)^{\frac{4}{5}}} \, dt \]

### Step 4: Simplify the Integral
Now we simplify the powers:

\[ 2 - \frac{6}{5} = \frac{10}{5} - \frac{6}{5} = \frac{4}{5} \]

Thus, the integral becomes:

\[ I = \frac{1}{2} \int t^{\frac{4}{5}} \, dt \]

### Step 5: Integrate
Now we can integrate:

\[ I = \frac{1}{2} \cdot \frac{t^{\frac{4}{5} + 1}}{\frac{4}{5} + 1} + C = \frac{1}{2} \cdot \frac{t^{\frac{9}{5}}}{\frac{9}{5}} + C = \frac{5}{18} t^{\frac{9}{5}} + C \]

### Step 6: Substitute Back
Substituting back for \(t\):

\[ t = \frac{x^2 + 1}{x^2 + 2} \]

Thus, we have:

\[ I = \frac{5}{18} \left(\frac{x^2 + 1}{x^2 + 2}\right)^{\frac{9}{5}} + C \]

### Final Answer
The final answer is:

\[ \int \frac{x \, dx}{(x^2 + 1)^{\frac{4}{5}} (x^2 + 2)^{\frac{6}{5}}} = \frac{5}{18} \left(\frac{x^2 + 1}{x^2 + 2}\right)^{\frac{9}{5}} + C \]