Integrate the following w.r.t x:
`(x^2+3x+1)/sqrt(1-x^2)`

Answer

To solve the integral

\[ \int \frac{x^2 + 3x + 1}{\sqrt{1 - x^2}} \, dx, \]

we will use the substitution \( x = \sin \theta \). This substitution simplifies the square root in the denominator.

### Step 1: Substitute \( x = \sin \theta \)

Differentiating both sides, we have:

\[ dx = \cos \theta \, d\theta. \]

Now, substituting \( x = \sin \theta \) into the integral:

\[ \sqrt{1 - x^2} = \sqrt{1 - \sin^2 \theta} = \sqrt{\cos^2 \theta} = \cos \theta. \]

Thus, the integral becomes:

\[ \int \frac{\sin^2 \theta + 3\sin \theta + 1}{\cos \theta} \cos \theta \, d\theta = \int (\sin^2 \theta + 3\sin \theta + 1) \, d\theta. \]

### Step 2: Simplify the Integral

Now we can break this integral into three separate integrals:

\[ \int \sin^2 \theta \, d\theta + 3 \int \sin \theta \, d\theta + \int 1 \, d\theta. \]

### Step 3: Integrate Each Term

1. **Integrate \( \sin^2 \theta \)**:

We use the identity \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \):

\[ \int \sin^2 \theta \, d\theta = \int \frac{1 - \cos 2\theta}{2} \, d\theta = \frac{1}{2} \left( \theta - \frac{1}{2} \sin 2\theta \right) + C_1. \]

2. **Integrate \( 3 \sin \theta \)**:

\[ 3 \int \sin \theta \, d\theta = -3 \cos \theta + C_2. \]

3. **Integrate \( 1 \)**:

\[ \int 1 \, d\theta = \theta + C_3. \]

### Step 4: Combine the Results

Combining all these results, we have:

\[ \int \sin^2 \theta \, d\theta + 3 \int \sin \theta \, d\theta + \int 1 \, d\theta = \frac{1}{2} \left( \theta - \frac{1}{2} \sin 2\theta \right) - 3 \cos \theta + \theta + C. \]

This simplifies to:

\[ \frac{3}{2} \theta - 3 \cos \theta - \frac{1}{4} \sin 2\theta + C. \]

### Step 5: Substitute Back to \( x \)

Recall that \( \theta = \sin^{-1}(x) \) and \( \cos \theta = \sqrt{1 - x^2} \). Also, \( \sin 2\theta = 2 \sin \theta \cos \theta = 2x\sqrt{1 - x^2} \).

Thus, substituting back, we get:

\[ \frac{3}{2} \sin^{-1}(x) - 3\sqrt{1 - x^2} - \frac{1}{4}(2x\sqrt{1 - x^2}) + C. \]

This simplifies to:

\[ \frac{3}{2} \sin^{-1}(x) - 3\sqrt{1 - x^2} - \frac{1}{2} x\sqrt{1 - x^2} + C. \]

### Final Answer

So the final result of the integral is:

\[ \int \frac{x^2 + 3x + 1}{\sqrt{1 - x^2}} \, dx = \frac{3}{2} \sin^{-1}(x) - 3\sqrt{1 - x^2} - \frac{1}{2} x\sqrt{1 - x^2} + C. \]