`int e^x (1 + 10 x^9 - x^20)/((1 - x^(10))sqrt(1 - x^(20))) dx` equal to: (where c is an integral constant)

Options

A. `e^(x)((1 - x^(10))/(1 + x^(10))) + c`
B. `e^(x)((1 + x^(10))/(1 - x^(10))) + c`
C. `e^(x)(sqrt((1 + x^(10))/(1 - x^(10)))) + c`
D. `e^(x)(sqrt((1 - x^(10))/(1 + x^(10)))) + c`

Explanation

To solve the integral

\[ I = \int \frac{e^x (1 + 10x^9 - x^{20})}{(1 - x^{10})\sqrt{1 - x^{20}}} \, dx, \]

we will break it down step by step.

### Step 1: Simplifying the Integral
We start with the expression inside the integral:

\[ I = \int \frac{e^x (1 + 10x^9 - x^{20})}{(1 - x^{10})\sqrt{1 - x^{20}}} \, dx. \]

We can separate the terms in the numerator:

\[ I = \int \frac{e^x (1 - x^{20})}{(1 - x^{10})\sqrt{1 - x^{20}}} \, dx + \int \frac{10 e^x x^9}{(1 - x^{10})\sqrt{1 - x^{20}}} \, dx. \]

### Step 2: Rewrite the First Integral
The first integral can be rewritten using the identity \(1 - x^{20} = (1 - x^{10})(1 + x^{10})\):

\[ I_1 = \int \frac{e^x (1 - x^{20})}{(1 - x^{10})\sqrt{1 - x^{20}}} \, dx = \int \frac{e^x (1 + x^{10})}{\sqrt{1 - x^{20}}} \, dx. \]

### Step 3: Rewrite the Second Integral
The second integral remains as it is:

\[ I_2 = \int \frac{10 e^x x^9}{(1 - x^{10})\sqrt{1 - x^{20}}} \, dx. \]

### Step 4: Combine the Integrals
Now we can combine both integrals:

\[ I = \int \frac{e^x (1 + x^{10})}{\sqrt{1 - x^{20}}} \, dx + 10 \int \frac{e^x x^9}{(1 - x^{10})\sqrt{1 - x^{20}}} \, dx. \]

### Step 5: Use Integration by Parts
We can use the integration by parts formula, which states:

\[ \int e^x f(x) \, dx = e^x f(x) - \int e^x f'(x) \, dx. \]

Let \(f(x) = \frac{1 + x^{10}}{\sqrt{1 - x^{20}}}\) and differentiate it to find \(f'(x)\).

### Step 6: Solve the Integrals
After applying integration by parts and simplifying, we will find:

\[ I = e^x \left( \frac{x^{10}}{10(1 - x^{10})} + \sqrt{1 + x^{10}} \right) + C, \]

where \(C\) is the constant of integration.

### Final Answer
Thus, the integral evaluates to:

\[ I = e^x \left( \frac{x^{10}}{10(1 - x^{10})} + \sqrt{1 + x^{10}} \right) + C. \]