Evaluate: `intx/((x^2+4)sqrt(x^2+1))dx`

Explanation

To evaluate the integral

\[ \int \frac{x}{(x^2 + 4) \sqrt{x^2 + 1}} \, dx, \]

we will use a substitution method. Let's follow the steps outlined in the video transcript.

### Step 1: Substitution
Let

\[ t = \sqrt{x^2 + 1}. \]

Then, we have

\[ t^2 = x^2 + 1 \implies x^2 = t^2 - 1. \]

Also, we can express \(x^2 + 4\) in terms of \(t\):

\[ x^2 + 4 = (t^2 - 1) + 4 = t^2 + 3. \]

### Step 2: Differentiate to find \(dx\)
Differentiating \(t\) with respect to \(x\):

\[ \frac{dt}{dx} = \frac{x}{\sqrt{x^2 + 1}} \implies dt = \frac{x}{\sqrt{x^2 + 1}} \, dx. \]

From this, we can express \(x \, dx\):

\[ x \, dx = t \, dt. \]

### Step 3: Substitute in the integral
Now, substituting \(x \, dx\), \(x^2 + 4\), and \(\sqrt{x^2 + 1}\) into the integral:

\[ \int \frac{x}{(x^2 + 4) \sqrt{x^2 + 1}} \, dx = \int \frac{t \, dt}{(t^2 + 3) t}. \]

This simplifies to:

\[ \int \frac{dt}{t^2 + 3}. \]

### Step 4: Integral of the simplified expression
Now we can evaluate the integral:

\[ \int \frac{dt}{t^2 + 3} = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) + C, \]

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

### Step 5: Substitute back for \(t\)
Recall that \(t = \sqrt{x^2 + 1}\). Thus, we substitute back:

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

### Final Result
The final result of the integral is:

\[ \int \frac{x}{(x^2 + 4) \sqrt{x^2 + 1}} \, dx = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{x^2 + 1}}{\sqrt{3}}\right) + C. \]

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