Evaluate: `int xsqrt(x^2+1) dx.`

Answer

To evaluate the integral \( \int x \sqrt{x^2 + 1} \, dx \), we can use substitution. Here is a step-by-step solution:

### Step 1: Choose a substitution
Let us set:
\[ t = \sqrt{x^2 + 1} \]
Then, squaring both sides gives:
\[ t^2 = x^2 + 1 \quad \Rightarrow \quad x^2 = t^2 - 1 \]

### Step 2: Differentiate the substitution
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(t^2) = \frac{d}{dx}(x^2 + 1) \quad \Rightarrow \quad 2t \frac{dt}{dx} = 2x \]
From this, we can solve for \( dx \):
\[ \frac{dt}{dx} = \frac{x}{t} \quad \Rightarrow \quad dx = \frac{t}{x} dt \]

### Step 3: Express \( x \) in terms of \( t \)
From our substitution \( t = \sqrt{x^2 + 1} \), we can express \( x \) as:
\[ x = \sqrt{t^2 - 1} \]

### Step 4: Substitute in the integral
Now we substitute \( x \) and \( dx \) into the integral:
\[ \int x \sqrt{x^2 + 1} \, dx = \int \sqrt{t^2 - 1} \cdot t \cdot \frac{t}{\sqrt{t^2 - 1}} dt \]
This simplifies to:
\[ \int t^2 \, dt \]

### Step 5: Integrate
Now we can integrate:
\[ \int t^2 \, dt = \frac{t^3}{3} + C \]

### Step 6: Substitute back
Finally, we substitute back \( t = \sqrt{x^2 + 1} \):
\[ \frac{(\sqrt{x^2 + 1})^3}{3} + C = \frac{(x^2 + 1)^{3/2}}{3} + C \]

### Final Answer
Thus, the evaluated integral is:
\[ \int x \sqrt{x^2 + 1} \, dx = \frac{(x^2 + 1)^{3/2}}{3} + C \]

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