The integral `int[2x^[12]+5x^9]/[x^5+x^3+1]^3.dx` is equal to-
(A) `x^10 / (2(x^5 + x^3 +1)^2) `
(B) `x^5/ (2(x^5 + x^3 +1)^2) `
(C) `-x^10 / (2(x^5 + x^3 +1)^2) `
(D) `- x^5 / (2(x^5 + x^3 +1)^2) `

Options

A. `(x^(10))/(2(x^(5)+x^(3)+1)^(2))+C`
B. `(x^(5))/(2(x^(5)+x^(3)+1)^(2))+C`
C. `(-x^(10))/(2(x^(5)+x^(3)+1)^(2))`
D. `(-x^(5))/((x^(5)+x^(3)+1)^(2))+C`

Explanation

To solve the integral

\[ I = \int \frac{2x^{12} + 5x^9}{(x^5 + x^3 + 1)^3} \, dx \]

we will follow these steps:

### Step 1: Divide the numerator and denominator by the highest power of \(x\)

The highest power in the denominator is \(x^5\) raised to the power of 3, which gives \(x^{15}\). Thus, we divide both the numerator and denominator by \(x^{15}\):

\[ I = \int \frac{2x^{12}/x^{15} + 5x^9/x^{15}}{(x^5/x^5 + x^3/x^5 + 1/x^5)^3} \, dx \]

This simplifies to:

\[ I = \int \frac{2/x^3 + 5/x^6}{(1 + 1/x^2 + 1/x^5)^3} \, dx \]

### Step 2: Substitute \(t\)

Let

\[ t = 1 + \frac{1}{x^2} + \frac{1}{x^5} \]

Now we need to find \(dt\):

\[ dt = -\frac{2}{x^3} \, dx - \frac{5}{x^6} \, dx \]

This implies:

\[ dx = -\frac{x^3}{2 + 5/x^3} \, dt \]

### Step 3: Rewrite the integral in terms of \(t\)

Substituting \(dx\) and the expression for \(t\) into the integral gives:

\[ I = \int \frac{-dt}{t^3} \]

### Step 4: Integrate

The integral of \(-\frac{1}{t^3}\) is:

\[ \int -t^{-3} \, dt = \frac{1}{2} t^{-2} + C = \frac{1}{2} \cdot \frac{1}{t^2} + C \]

### Step 5: Substitute back for \(t\)

Substituting back for \(t\):

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

### Step 6: Simplify

This can be rewritten as:

\[ I = \frac{1}{2} \cdot \frac{1}{\left(x^5 + x^3 + 1\right)^2} \cdot x^{10} \]

### Final Result

Thus, the final answer is:

\[ I = \frac{x^{10}}{2(x^5 + x^3 + 1)^2} + C \]

### Conclusion

The correct option is (A)

\[ \frac{x^{10}}{2(x^5 + x^3 + 1)^2} \]