`int (x ^(n-1))/( x ^(2n) + a ^(2)) dx =`

Options

A. `(1)/(na) tan ^(-1) ((x ^(n))/( a )) +C`
B. `(n)/(a) tan ^(-1) ((x ^(n))/( a )) +C`
C. `(n)/(a) sin ^(-1) ((x ^(n))/(a)) +C`
D. `(n)/(a) cos ^(-1) ((x ^(n))/( a )) +C `

Answer

To solve the integral

\[ \int \frac{x^{n-1}}{x^{2n} + a^2} \, dx, \]

we will follow these steps:

### Step 1: Substitution
Let \( t = x^n \). Then, differentiating both sides gives:

\[ dt = n x^{n-1} \, dx \quad \Rightarrow \quad dx = \frac{dt}{n x^{n-1}}. \]

### Step 2: Express \( x^{n-1} \) in terms of \( t \)
From the substitution \( t = x^n \), we have:

\[ x^{n-1} = \frac{t^{(n-1)/n}}{x} = \frac{t^{(n-1)/n}}{t^{1/n}} = t^{(n-1)/n} \cdot t^{-1/n} = t^{(n-1)/n - 1/n} = t^{(n-2)/n}. \]

### Step 3: Substitute in the integral
Now we can rewrite the integral:

\[ \int \frac{x^{n-1}}{x^{2n} + a^2} \, dx = \int \frac{t^{(n-2)/n}}{t^2 + a^2} \cdot \frac{dt}{n t^{(n-1)/n}}. \]

This simplifies to:

\[ \frac{1}{n} \int \frac{dt}{t^2 + a^2}. \]

### Step 4: Use the formula for the integral
We know that:

\[ \int \frac{1}{t^2 + a^2} \, dt = \frac{1}{a} \tan^{-1}\left(\frac{t}{a}\right) + C. \]

Thus, we substitute this result into our integral:

\[ \frac{1}{n} \cdot \frac{1}{a} \tan^{-1}\left(\frac{t}{a}\right) + C. \]

### Step 5: Substitute back for \( t \)
Recalling that \( t = x^n \), we substitute back:

\[ \frac{1}{na} \tan^{-1}\left(\frac{x^n}{a}\right) + C. \]

### Final Answer
Thus, the final result for the integral is:

\[ \int \frac{x^{n-1}}{x^{2n} + a^2} \, dx = \frac{1}{na} \tan^{-1}\left(\frac{x^n}{a}\right) + C. \]

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