The value of ` int _(0)^(1) (x^(4) +1)/(x^(2)+1)dx ` is

Options

A. `1/6(3-4pi)`
B. `1/6 (3pi+4)`
C. `1/6(3+4pi)`
D. `1/6(3pi-4)`

Explanation

To solve the integral \( \int_0^1 \frac{x^4 + 1}{x^2 + 1} \, dx \), we can break it down step by step.

### Step 1: Rewrite the integrand
We start by rewriting the integrand:
\[ \frac{x^4 + 1}{x^2 + 1} = \frac{x^4 - 1 + 2}{x^2 + 1} = \frac{(x^4 - 1) + 2}{x^2 + 1} \]
This allows us to separate the integral into two parts:
\[ \int_0^1 \frac{x^4 + 1}{x^2 + 1} \, dx = \int_0^1 \frac{x^4 - 1}{x^2 + 1} \, dx + \int_0^1 \frac{2}{x^2 + 1} \, dx \]

### Step 2: Simplify the first integral
Next, we simplify the first integral:
\[ \int_0^1 \frac{x^4 - 1}{x^2 + 1} \, dx = \int_0^1 \frac{(x^2 - 1)(x^2 + 1)}{x^2 + 1} \, dx = \int_0^1 (x^2 - 1) \, dx \]
The \(x^2 + 1\) terms cancel out.

### Step 3: Evaluate the first integral
Now we evaluate the integral:
\[ \int_0^1 (x^2 - 1) \, dx = \int_0^1 x^2 \, dx - \int_0^1 1 \, dx \]
Calculating each part:
\[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \]
\[ \int_0^1 1 \, dx = [x]_0^1 = 1 \]
Thus,
\[ \int_0^1 (x^2 - 1) \, dx = \frac{1}{3} - 1 = -\frac{2}{3} \]

### Step 4: Evaluate the second integral
Now we evaluate the second integral:
\[ \int_0^1 \frac{2}{x^2 + 1} \, dx = 2 \int_0^1 \frac{1}{x^2 + 1} \, dx \]
The integral \( \int \frac{1}{x^2 + 1} \, dx \) is known to be \( \tan^{-1}(x) \):
\[ \int_0^1 \frac{1}{x^2 + 1} \, dx = \left[ \tan^{-1}(x) \right]_0^1 = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]
Thus,
\[ \int_0^1 \frac{2}{x^2 + 1} \, dx = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} \]

### Step 5: Combine results
Now we can combine the results from the two integrals:
\[ \int_0^1 \frac{x^4 + 1}{x^2 + 1} \, dx = -\frac{2}{3} + \frac{\pi}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_0^1 \frac{x^4 + 1}{x^2 + 1} \, dx = \frac{\pi}{2} - \frac{2}{3} \]